Hello- the following code- In LeftRotaate and RightRotate methods for.pdf
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The document is a request for help fixing an error in some code related to inserting nodes into a binary search tree. Specifically, the methods LeftRotaate and RightRotate are causing errors when referencing 'parent', indicating it is not defined. The request is for help correcting this issue with the 'parent' reference.
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Assignment 9 (Parent reference for BST) Redefine TreeNode by adding .pdf
Assignment 9 (Parent reference for BST) Redefine TreeNode by adding a reference to a node’s
parent, as shown below:
BST.TreeNode
#element
#left : TreeNode
#right: TreeNode
#parent: TreeNode
Implement the insert and delete methods in BST class to update the parent for each node in the
tree.
Add the following new method in BST:
/** Return the node for the specific element.
* Return null if the element is not in the tree. */
private TreeNode getNode(E element)
/** Returns true if the node for the element is a lead */
private boolean isLeaf(E element)
/** Returns the path of the element from the specific element to the root in an array list */
public ArrayList getPath(E e)
Write a test program that prompts the user to enter 10 integers, adds them to the tree, deletes the
first integer from the tree, and displays the paths for all leaf nodes.
Sample run
Enter 10 integers :
45 54 67 56 50 45 23 59 23 67
[50, 54, 23]
[59, 56, 67, 54, 23]
Solution
import java.util.NoSuchElementException; public class TreeSet> implements Set {
private TreeNode root; private int size; // Constructs an empty set. public
TreeSet() { root = null; size = 0; } // Adds the given value to this
set, if it was not already in the set. // If the element is already a member of this set, adding it
again has no effect. public void add(E value) { root = add(root, value); }
// private recursive helper for add // uses the \"x = change(x)\" pattern: // pass in
current state of root // return out desired new state of root private TreeNode
add(TreeNode node, E value) { if (node == null) { node = new
TreeNode(value); // reached dead end; put new node here size++; }
else if (value.compareTo(node.data) > 0) { node.right = add(node.right, value);
} else if (value.compareTo(node.data) < 0) { node.left = add(node.left,
value); } // else a duplicate; do nothing return node; } // Returns
true if this set contains the given element value. public boolean contains(E value) {
return contains(root, value); } // private recursive helper for contains private
boolean contains(TreeNode node, E value) { if (node == null) { return
false; } else if (value.compareTo(node.data) > 0) { return
contains(node.right, value); } else if (value.compareTo(node.data) < 0) {
return contains(node.left, value); } else { // value.equals(root.data);
found it! return true; } } // Returns the minimum value in this
set. // Throws a NoSuchElementException if this set is empty. public E getMin() {
if (root == null) { throw new NoSuchElementException(\"empty tree\");
} return getMin(root); } // private recursive helper for getMin
private E getMin(TreeNode node) { if (node.left == null) { return
node.data; } else { return getMin(node.left); } }
// Returns true if this set does not contain any elements. public boolean isEmpty() {
return size == 0; } // Prints all elements of this tree in a left-to-right order (in-order)
traversal. public void print() { print(root); } // private recursive helper
for print private void print(TreeNode .
Extend your work on the OCC logo to add shapes that curve1 the blue.pdf
Extend your work on the OCC logo to add shapes that \'curve1 the blue Cs into looking
something more like the actual logo:
Solution
import java.awt.Color;
import java.util.Comparator;
/**
* a straightforward red-black tree category.
*/
public category RedBlackTree extends BinarySearchTree Associate in Nursing empty
RedBlackTree which will solely settle for Comparables as
* items.
*/
public RedBlackTree()
/**
* Constructs Associate in Nursing empty RedBlackTree that orders its things in keeping with
the
* given comparator.
*/
public RedBlackTree(Comparator c)
/**
* The nodes during a red-black tree store a color beside the particular information
* within the node.
*/
category Node extends LinkedBinaryTreeNode
}
/**
* Adds one information item to the tree. If there\'s already Associate in Nursing item within the
* tree that compares up to the item being inserted, it\'s \"overwritten\"
* by the new item. Overrides BinarySearchTree.add as a result of the tree must
* be adjusted when insertion.
*/
public void add(Object data)
BinaryTreeNode n = root;
whereas (true) else if (comparisonResult < 0)
n = n.getLeft();
} else zero
if (n.getRight() == null)
n = n.getRight();
}
}
}
/**
* Removes the node containing the given price. will nothing if there\'s no
* such node.
*/
public void remove(Object data) else if (node.getLeft() != null && node.getRight() != null) 2
kids, Copy forerunner information in.
BinaryTreeNode forerunner = predecessor(node);
node.setData(predecessor.getData());
node = (Node) predecessor;
}
// At this time node has zero or one kid
Node pullUp = leftOf(node) == null ? rightOf(node) : leftOf(node);
if (pullUp != null) regulate if pullUp could be a double black.
if (node == root) else if (node.getParent().getLeft() == node) else
if (isBlack(node))
} else if (node == root) to drag up once deleting a root means that we have a tendency to empty
the tree
setRoot(null);
} else lookout
if (isBlack(node))
node.removeFromParent();
}
}
/**
* Classic formula for fixing up a tree when inserting a node.
*/
personal void adjustAfterInsertion(Node n) issues, if they exist
if (n != null && n != root && isRed(parentOf(n))) to ascertain if additional work
// needed
if (isRed(siblingOf(parentOf(n))))
// Step 2b: reconstitute for a parent United Nations agency is that the left kid of the
// forebear. this may need one right rotation if n is
// also
// a left kid, or a left-right rotation otherwise.
else if (parentOf(n) == leftOf(grandparentOf(n)))
setColor(parentOf(n), Color.black);
setColor(grandparentOf(n), Color.red);
rotateRight(grandparentOf(n));
}
// Step 2c: reconstitute for a parent United Nations agency is that the right kid of the
// forebear. this may need one left rotation if n is
// also
// a right kid, or a right-left rotation otherwise.
else if (parentOf(n) == rightOf(grandparentOf(n)))
setColor(parentOf(n), Color.black);
setColor(grandparentOf(n), Color.red);
rotateLeft(grandparentOf(n));
}
}
// Step 3: Color the foundation black
setColo.
Binary search tree exact match - illustrated walkthrough
A step-by-step illustration of Binary Search Tree (Exact Match) to help you walk through a series of operations. Illustration is accompanied by actual code with bold line indicating the current operation.
fix the error - class Node{ int data- Node next-.pdf
fix the error :
class Node{
int data;
Node next;
public Node(int data) {
this.data = data;
this.next = null;
}
}
class singlelinkedlist {
public Node head = null;
public Node tail = null;
public void addNode(int data) {
Node newNode = new Node(data);
if(head == null) {
head = newNode;
tail = newNode;
}
else {
tail.next = newNode;
tail = newNode;
}
}
public void sortList() {
Node current = head, index = null;
int temp;
if(head == null) {
return;
}
else {
while(current != null) {
index = current.next;
while(index != null) {
if(current.data > index.data) {
temp = current.data;
current.data = index.data;
index.data = temp;
}
index = index.next;
}
current = current.next;
}
}
}
public void display() {
Node current = head;
if(head == null) {
System.out.println("List is empty");
return;
}
while(current != null) {
System.out.print(current.data + " ");
current = current.next;
}
System.out.println();
}
}
public class JavaApplication46 {
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
// TODO code application logic here
SortList sList = new SortList();
sList.addNode(9);
sList.addNode(7);
sList.addNode(2);
sList.addNode(5);
sList.addNode(4);
System.out.println("Original list: ");
sList.display();
sList.sortList();
System.out.println("Sorted list: ");
sList.display();
}
}.
I need help finishing this code in JavaYou will need to create t.pdf
I need help finishing this code in Java
You will need to create the preOrder and postOrder traversal methods, as these will be used to
determine if your insertion methods are correct.
protected class Node {
public int data;
public Node left, right, parent;
public Node(int _data) {
data = _data;
left = right = parent = null;
}
public Node(int _data, Node _parent) {
data = _data;
left = right = null;
parent = _parent;
}
}
protected Node root;
public BST() {
root = null;
}
public void insert(int data) {
if (root == null) {
root = new Node(data);
} else {
insert(data, root);
}
}
protected void insert(int data, Node n) {
// TODO: Insert Method
}
public ArrayList inOrderTraversal() {
return inOrderTraversal(root);
}
private ArrayList inOrderTraversal(Node n) {
if (n == null) {
return new ArrayList<>();
}
ArrayList traversal = new ArrayList<>();
traversal.addAll(inOrderTraversal(n.left));
traversal.add(n.data);
traversal.addAll(inOrderTraversal(n.right));
return traversal;
}
public ArrayList preOrderTraversal() {
return preOrderTraversal(root);
}
private ArrayList preOrderTraversal(Node n) {
// TODO: preOrderTraveral Method
return new ArrayList();
}
public ArrayList postOrderTraversal() {
return postOrderTraversal(root);
}
private ArrayList postOrderTraversal(Node n) {
// TODO: postOrderTraversal Method
return new ArrayList();
}
protected static int height(Node n) {
if (n == null) {
return -1;
}
int lheight = height(n.left);
int rheight = height(n.right);
if (lheight > rheight) {
return 1 + lheight;
}
return 1 + rheight;
}
public int height() {
return height(root);
}
} BST Insertion protected void insert(int data, Node n ) if data < n.data if n. left is null n.left =
new Node(data, n ) else insert(data, n.left) else if data > n.data if n.right is null n.right = new
Node(data, n) else insert(data, n.right)
BST Traversals You are responsible for implementing the and postOrderTraversal(Node n )
methods that generate the string representation of the tree using the specified traversal. The
inorderTraversal (Node n ) method has been implemented for you; use that to match the
formatting of the output strings (this is the most "output matching" you will have to do this
semester)..
Count Red Nodes. Write a program that computes the percentage of red.pdf
Count Red Nodes. Write a program that computes the percentage of red nodes in a given red-
black BST. Test your program by running at least 100 trials of the experiment of inserting N
random keys into an initially empty tree, for N = 10^4, 10^5, 10^6, and formulate an hypothesis.
*It is experiment from textbook Algorithms, 3.3.42
Does anyone know the answer?
Solution
/*
* Java Program to Implement Red Black Tree
*/
import java.util.Scanner;
/* Class Node */
class RedBlackNode
{
RedBlackNode left, right;
int element;
int color;
/* Constructor */
public RedBlackNode(int theElement)
{
this( theElement, null, null );
}
/* Constructor */
public RedBlackNode(int theElement, RedBlackNode lt, RedBlackNode rt)
{
left = lt;
right = rt;
element = theElement;
color = 1;
}
}
/* Class RBTree */
class RBTree
{
private RedBlackNode current;
private RedBlackNode parent;
private RedBlackNode grand;
private RedBlackNode great;
private RedBlackNode header;
private static RedBlackNode nullNode;
/* static initializer for nullNode */
static
{
nullNode = new RedBlackNode(0);
nullNode.left = nullNode;
nullNode.right = nullNode;
}
/* Black - 1 RED - 0 */
static final int BLACK = 1;
static final int RED = 0;
/* Constructor */
public RBTree(int negInf)
{
header = new RedBlackNode(negInf);
header.left = nullNode;
header.right = nullNode;
}
/* Function to check if tree is empty */
public boolean isEmpty()
{
return header.right == nullNode;
}
/* Make the tree logically empty */
public void makeEmpty()
{
header.right = nullNode;
}
/* Function to insert item */
public void insert(int item )
{
current = parent = grand = header;
nullNode.element = item;
while (current.element != item)
{
great = grand;
grand = parent;
parent = current;
current = item < current.element ? current.left : current.right;
// Check if two red children and fix if so
if (current.left.color == RED && current.right.color == RED)
handleReorient( item );
}
// Insertion fails if already present
if (current != nullNode)
return;
current = new RedBlackNode(item, nullNode, nullNode);
// Attach to parent
if (item < parent.element)
parent.left = current;
else
parent.right = current;
handleReorient( item );
}
private void handleReorient(int item)
{
// Do the color flip
current.color = RED;
current.left.color = BLACK;
current.right.color = BLACK;
if (parent.color == RED)
{
// Have to rotate
grand.color = RED;
if (item < grand.element != item < parent.element)
parent = rotate( item, grand ); // Start dbl rotate
current = rotate(item, great );
current.color = BLACK;
}
// Make root black
header.right.color = BLACK;
}
private RedBlackNode rotate(int item, RedBlackNode parent)
{
if(item < parent.element)
return parent.left = item < parent.left.element ? rotateWithLeftChild(parent.left) :
rotateWithRightChild(parent.left) ;
else
return parent.right = item < parent.right.element ? rotateWithLeftChild(parent.right) :
rotateWithRightChild(parent.right);
}
/* Rotate binary tree node with left child */
private RedBlackNode rot.
Given a newly created Binary Search Tree with the following numerica.pdf
Given a newly created Binary Search Tree with the following numerical key input sequence
(from left to right): 9, 4, 12, 7, 5, 2, 20, 14, 11, 13, 19, 16, 17, 42, 24
a.) We are asked to add a function Position lowestKey(const Position& v) const to the class
SearchTree.
Function prototype: Position lowestKey(const Position& v);
input argument: position of the starting node in the binary search tree to calculate from. For the
whole tree this would be the position for the root of the tree.
return value: position of the node having the lowest key value.
i. Describe strategy with reasonings and examples to convince that it will work.
ii. Implement the function using C++ and show the source code listing.
b) We are also asked to add a function Position highestKey(const Position& v) const to the class
SearchTree.
Function prototype: Position highestKey(const Position& v) const;
input argument: position of the node in the binary search tree to calculate from. For the whole
tree this would be the position for the root of the tree.
return value: position of the node having the highest key value.
i. Describe strategy with reasonings and examples to convince that it will work.
ii. Implement the function using C++ and show the source code listing.
Solution
#include
#include
/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct node
{
int data;
struct node* left;
struct node* right;
};
/* Helper function that allocates a new node
with the given data and NULL left and right
pointers. */
struct node* newNode(int data)
{
struct node* node = (struct node*)
malloc(sizeof(struct node));
node->data = data;
node->left = NULL;
node->right = NULL;
return(node);
}
/* Give a binary search tree and a number,
inserts a new node with the given number in
the correct place in the tree. Returns the new
root pointer which the caller should then use
(the standard trick to avoid using reference
parameters). */
struct node* insert(struct node* node, int data)
{
/* 1. If the tree is empty, return a new,
single node */
if (node == NULL)
return(newNode(data));
else
{
/* 2. Otherwise, recur down the tree */
if (data <= node->data)
node->left = insert(node->left, data);
else
node->right = insert(node->right, data);
/* return the (unchanged) node pointer */
return node;
}
}
/* Given a non-empty binary search tree,
return the minimum data value found in that
tree. Note that the entire tree does not need
to be searched. */
int minValue(struct node* node) {
struct node* current = node;
/* loop down to find the leftmost leaf */
while (current->left != NULL) {
current = current->left;
}
return(current->data);
}
/* Driver program to test sameTree function*/
int main()
{
struct node* root = NULL;
root = insert(root, 4);
insert(root, 2);
insert(root, 1);
insert(root, 3);
insert(root, 6);
insert(root, 5);
printf(\"\ Minimum value in BST is %d\", minValue(root));
getchar();
return 0;
}
#include
#include
/* A binary tree node has data, pointer to left child
.
You can list anything, it doesnt matter. I just want to see code f.pdf
You can list anything, it doesn\'t matter. I just want to see code for this problem. Thank you
Solution
Include header files .......
Source code:
public class SinglyLinkedListImpl {
private Node head;
private Node tail;
public void add(T element){
Node nd = new Node();
nd.setValue(element);
System.out.println(\"Adding: \"+element);
if(head == null){
head = nd;
tail = nd;
} else {
tail.setNextRef(nd);
}
tail = nd;
}
}
public void addAfter(T element, T after){
Node tmp = head;
Node refNode = null;
System.out.println(\"Traversing to all nodes..\");
while(true){
if(tmp == null){
break;
}
if(tmp.compareTo(after) == 0){
refNode = tmp;
break;
}
tmp = tmp.getNextRef();
}
if(refNode != null){
Node nd = new Node();
nd.setValue(element);
nd.setNextRef(tmp.getNextRef());
if(tmp == tail){
tail = nd;
}
tmp.setNextRef(nd);
} else {
System.out.println(\"Unable to find the given element...\");
}
}
public void deleteFront(){
if(head == null){
System.out.println(\"Underflow...\");
}
Node tmp = head;
head = tmp.getNextRef();
if(head == null){
tail = null;
}
System.out.println(\"Deleted: \"+tmp.getValue());
}
public void deleteAfter(T after){
Node tmp = head;
Node refNode = null;
System.out.println(\"Traversing to all nodes..\");
while(true){
if(tmp == null){
break;
}
if(tmp.compareTo(after) == 0){
refNode = tmp;
break;
}
tmp = tmp.getNextRef();
}
if(refNode != null){
tmp = refNode.getNextRef();
refNode.setNextRef(tmp.getNextRef());
if(refNode.getNextRef() == null){
tail = refNode;
}
System.out.println(\"Deleted: \"+tmp.getValue());
} else {
System.out.println(\"Unable to find the given element...\");
}
}
public void traverse(){
Node tmp = head;
while(true){
if(tmp == null){
break;
}
System.out.println(tmp.getValue());
tmp = tmp.getNextRef();
}
}
public static void main(String a[]){
SinglyLinkedListImpl sl = new SinglyLinkedListImpl();
sl.add(3);
sl.add(32);
sl.add(54);
sl.add(89);
sl.addAfter(76, 54);
sl.deleteFront();
sl.deleteAfter(76);
sl.traverse();
}
}
class Node implements Comparable {
private T value;
private Node nextRef;
public T getValue() {
return value;
}
public void setValue(T value) {
this.value = value;
}
public Node getNextRef() {
return nextRef;
}
public void setNextRef(Node ref) {
this.nextRef = ref;
}
public int compareTo(T arg) {
if(arg == this.value){
return 0;
} else {
return 1;
}
}
}
This program implement the concept of single linked list..
Recommended
Assignment 9 (Parent reference for BST) Redefine TreeNode by adding .pdf
Assignment 9 (Parent reference for BST) Redefine TreeNode by adding a reference to a node’s
parent, as shown below:
BST.TreeNode
#element
#left : TreeNode
#right: TreeNode
#parent: TreeNode
Implement the insert and delete methods in BST class to update the parent for each node in the
tree.
Add the following new method in BST:
/** Return the node for the specific element.
* Return null if the element is not in the tree. */
private TreeNode getNode(E element)
/** Returns true if the node for the element is a lead */
private boolean isLeaf(E element)
/** Returns the path of the element from the specific element to the root in an array list */
public ArrayList getPath(E e)
Write a test program that prompts the user to enter 10 integers, adds them to the tree, deletes the
first integer from the tree, and displays the paths for all leaf nodes.
Sample run
Enter 10 integers :
45 54 67 56 50 45 23 59 23 67
[50, 54, 23]
[59, 56, 67, 54, 23]
Solution
import java.util.NoSuchElementException; public class TreeSet> implements Set {
private TreeNode root; private int size; // Constructs an empty set. public
TreeSet() { root = null; size = 0; } // Adds the given value to this
set, if it was not already in the set. // If the element is already a member of this set, adding it
again has no effect. public void add(E value) { root = add(root, value); }
// private recursive helper for add // uses the \"x = change(x)\" pattern: // pass in
current state of root // return out desired new state of root private TreeNode
add(TreeNode node, E value) { if (node == null) { node = new
TreeNode(value); // reached dead end; put new node here size++; }
else if (value.compareTo(node.data) > 0) { node.right = add(node.right, value);
} else if (value.compareTo(node.data) < 0) { node.left = add(node.left,
value); } // else a duplicate; do nothing return node; } // Returns
true if this set contains the given element value. public boolean contains(E value) {
return contains(root, value); } // private recursive helper for contains private
boolean contains(TreeNode node, E value) { if (node == null) { return
false; } else if (value.compareTo(node.data) > 0) { return
contains(node.right, value); } else if (value.compareTo(node.data) < 0) {
return contains(node.left, value); } else { // value.equals(root.data);
found it! return true; } } // Returns the minimum value in this
set. // Throws a NoSuchElementException if this set is empty. public E getMin() {
if (root == null) { throw new NoSuchElementException(\"empty tree\");
} return getMin(root); } // private recursive helper for getMin
private E getMin(TreeNode node) { if (node.left == null) { return
node.data; } else { return getMin(node.left); } }
// Returns true if this set does not contain any elements. public boolean isEmpty() {
return size == 0; } // Prints all elements of this tree in a left-to-right order (in-order)
traversal. public void print() { print(root); } // private recursive helper
for print private void print(TreeNode .
Extend your work on the OCC logo to add shapes that curve1 the blue.pdf
Extend your work on the OCC logo to add shapes that \'curve1 the blue Cs into looking
something more like the actual logo:
Solution
import java.awt.Color;
import java.util.Comparator;
/**
* a straightforward red-black tree category.
*/
public category RedBlackTree extends BinarySearchTree Associate in Nursing empty
RedBlackTree which will solely settle for Comparables as
* items.
*/
public RedBlackTree()
/**
* Constructs Associate in Nursing empty RedBlackTree that orders its things in keeping with
the
* given comparator.
*/
public RedBlackTree(Comparator c)
/**
* The nodes during a red-black tree store a color beside the particular information
* within the node.
*/
category Node extends LinkedBinaryTreeNode
}
/**
* Adds one information item to the tree. If there\'s already Associate in Nursing item within the
* tree that compares up to the item being inserted, it\'s \"overwritten\"
* by the new item. Overrides BinarySearchTree.add as a result of the tree must
* be adjusted when insertion.
*/
public void add(Object data)
BinaryTreeNode n = root;
whereas (true) else if (comparisonResult < 0)
n = n.getLeft();
} else zero
if (n.getRight() == null)
n = n.getRight();
}
}
}
/**
* Removes the node containing the given price. will nothing if there\'s no
* such node.
*/
public void remove(Object data) else if (node.getLeft() != null && node.getRight() != null) 2
kids, Copy forerunner information in.
BinaryTreeNode forerunner = predecessor(node);
node.setData(predecessor.getData());
node = (Node) predecessor;
}
// At this time node has zero or one kid
Node pullUp = leftOf(node) == null ? rightOf(node) : leftOf(node);
if (pullUp != null) regulate if pullUp could be a double black.
if (node == root) else if (node.getParent().getLeft() == node) else
if (isBlack(node))
} else if (node == root) to drag up once deleting a root means that we have a tendency to empty
the tree
setRoot(null);
} else lookout
if (isBlack(node))
node.removeFromParent();
}
}
/**
* Classic formula for fixing up a tree when inserting a node.
*/
personal void adjustAfterInsertion(Node n) issues, if they exist
if (n != null && n != root && isRed(parentOf(n))) to ascertain if additional work
// needed
if (isRed(siblingOf(parentOf(n))))
// Step 2b: reconstitute for a parent United Nations agency is that the left kid of the
// forebear. this may need one right rotation if n is
// also
// a left kid, or a left-right rotation otherwise.
else if (parentOf(n) == leftOf(grandparentOf(n)))
setColor(parentOf(n), Color.black);
setColor(grandparentOf(n), Color.red);
rotateRight(grandparentOf(n));
}
// Step 2c: reconstitute for a parent United Nations agency is that the right kid of the
// forebear. this may need one left rotation if n is
// also
// a right kid, or a right-left rotation otherwise.
else if (parentOf(n) == rightOf(grandparentOf(n)))
setColor(parentOf(n), Color.black);
setColor(grandparentOf(n), Color.red);
rotateLeft(grandparentOf(n));
}
}
// Step 3: Color the foundation black
setColo.
Binary search tree exact match - illustrated walkthrough
A step-by-step illustration of Binary Search Tree (Exact Match) to help you walk through a series of operations. Illustration is accompanied by actual code with bold line indicating the current operation.
fix the error - class Node{ int data- Node next-.pdf
fix the error :
class Node{
int data;
Node next;
public Node(int data) {
this.data = data;
this.next = null;
}
}
class singlelinkedlist {
public Node head = null;
public Node tail = null;
public void addNode(int data) {
Node newNode = new Node(data);
if(head == null) {
head = newNode;
tail = newNode;
}
else {
tail.next = newNode;
tail = newNode;
}
}
public void sortList() {
Node current = head, index = null;
int temp;
if(head == null) {
return;
}
else {
while(current != null) {
index = current.next;
while(index != null) {
if(current.data > index.data) {
temp = current.data;
current.data = index.data;
index.data = temp;
}
index = index.next;
}
current = current.next;
}
}
}
public void display() {
Node current = head;
if(head == null) {
System.out.println("List is empty");
return;
}
while(current != null) {
System.out.print(current.data + " ");
current = current.next;
}
System.out.println();
}
}
public class JavaApplication46 {
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
// TODO code application logic here
SortList sList = new SortList();
sList.addNode(9);
sList.addNode(7);
sList.addNode(2);
sList.addNode(5);
sList.addNode(4);
System.out.println("Original list: ");
sList.display();
sList.sortList();
System.out.println("Sorted list: ");
sList.display();
}
}.
I need help finishing this code in JavaYou will need to create t.pdf
I need help finishing this code in Java
You will need to create the preOrder and postOrder traversal methods, as these will be used to
determine if your insertion methods are correct.
protected class Node {
public int data;
public Node left, right, parent;
public Node(int _data) {
data = _data;
left = right = parent = null;
}
public Node(int _data, Node _parent) {
data = _data;
left = right = null;
parent = _parent;
}
}
protected Node root;
public BST() {
root = null;
}
public void insert(int data) {
if (root == null) {
root = new Node(data);
} else {
insert(data, root);
}
}
protected void insert(int data, Node n) {
// TODO: Insert Method
}
public ArrayList inOrderTraversal() {
return inOrderTraversal(root);
}
private ArrayList inOrderTraversal(Node n) {
if (n == null) {
return new ArrayList<>();
}
ArrayList traversal = new ArrayList<>();
traversal.addAll(inOrderTraversal(n.left));
traversal.add(n.data);
traversal.addAll(inOrderTraversal(n.right));
return traversal;
}
public ArrayList preOrderTraversal() {
return preOrderTraversal(root);
}
private ArrayList preOrderTraversal(Node n) {
// TODO: preOrderTraveral Method
return new ArrayList();
}
public ArrayList postOrderTraversal() {
return postOrderTraversal(root);
}
private ArrayList postOrderTraversal(Node n) {
// TODO: postOrderTraversal Method
return new ArrayList();
}
protected static int height(Node n) {
if (n == null) {
return -1;
}
int lheight = height(n.left);
int rheight = height(n.right);
if (lheight > rheight) {
return 1 + lheight;
}
return 1 + rheight;
}
public int height() {
return height(root);
}
} BST Insertion protected void insert(int data, Node n ) if data < n.data if n. left is null n.left =
new Node(data, n ) else insert(data, n.left) else if data > n.data if n.right is null n.right = new
Node(data, n) else insert(data, n.right)
BST Traversals You are responsible for implementing the and postOrderTraversal(Node n )
methods that generate the string representation of the tree using the specified traversal. The
inorderTraversal (Node n ) method has been implemented for you; use that to match the
formatting of the output strings (this is the most "output matching" you will have to do this
semester)..
Count Red Nodes. Write a program that computes the percentage of red.pdf
Count Red Nodes. Write a program that computes the percentage of red nodes in a given red-
black BST. Test your program by running at least 100 trials of the experiment of inserting N
random keys into an initially empty tree, for N = 10^4, 10^5, 10^6, and formulate an hypothesis.
*It is experiment from textbook Algorithms, 3.3.42
Does anyone know the answer?
Solution
/*
* Java Program to Implement Red Black Tree
*/
import java.util.Scanner;
/* Class Node */
class RedBlackNode
{
RedBlackNode left, right;
int element;
int color;
/* Constructor */
public RedBlackNode(int theElement)
{
this( theElement, null, null );
}
/* Constructor */
public RedBlackNode(int theElement, RedBlackNode lt, RedBlackNode rt)
{
left = lt;
right = rt;
element = theElement;
color = 1;
}
}
/* Class RBTree */
class RBTree
{
private RedBlackNode current;
private RedBlackNode parent;
private RedBlackNode grand;
private RedBlackNode great;
private RedBlackNode header;
private static RedBlackNode nullNode;
/* static initializer for nullNode */
static
{
nullNode = new RedBlackNode(0);
nullNode.left = nullNode;
nullNode.right = nullNode;
}
/* Black - 1 RED - 0 */
static final int BLACK = 1;
static final int RED = 0;
/* Constructor */
public RBTree(int negInf)
{
header = new RedBlackNode(negInf);
header.left = nullNode;
header.right = nullNode;
}
/* Function to check if tree is empty */
public boolean isEmpty()
{
return header.right == nullNode;
}
/* Make the tree logically empty */
public void makeEmpty()
{
header.right = nullNode;
}
/* Function to insert item */
public void insert(int item )
{
current = parent = grand = header;
nullNode.element = item;
while (current.element != item)
{
great = grand;
grand = parent;
parent = current;
current = item < current.element ? current.left : current.right;
// Check if two red children and fix if so
if (current.left.color == RED && current.right.color == RED)
handleReorient( item );
}
// Insertion fails if already present
if (current != nullNode)
return;
current = new RedBlackNode(item, nullNode, nullNode);
// Attach to parent
if (item < parent.element)
parent.left = current;
else
parent.right = current;
handleReorient( item );
}
private void handleReorient(int item)
{
// Do the color flip
current.color = RED;
current.left.color = BLACK;
current.right.color = BLACK;
if (parent.color == RED)
{
// Have to rotate
grand.color = RED;
if (item < grand.element != item < parent.element)
parent = rotate( item, grand ); // Start dbl rotate
current = rotate(item, great );
current.color = BLACK;
}
// Make root black
header.right.color = BLACK;
}
private RedBlackNode rotate(int item, RedBlackNode parent)
{
if(item < parent.element)
return parent.left = item < parent.left.element ? rotateWithLeftChild(parent.left) :
rotateWithRightChild(parent.left) ;
else
return parent.right = item < parent.right.element ? rotateWithLeftChild(parent.right) :
rotateWithRightChild(parent.right);
}
/* Rotate binary tree node with left child */
private RedBlackNode rot.
Given a newly created Binary Search Tree with the following numerica.pdf
Given a newly created Binary Search Tree with the following numerical key input sequence
(from left to right): 9, 4, 12, 7, 5, 2, 20, 14, 11, 13, 19, 16, 17, 42, 24
a.) We are asked to add a function Position lowestKey(const Position& v) const to the class
SearchTree.
Function prototype: Position lowestKey(const Position& v);
input argument: position of the starting node in the binary search tree to calculate from. For the
whole tree this would be the position for the root of the tree.
return value: position of the node having the lowest key value.
i. Describe strategy with reasonings and examples to convince that it will work.
ii. Implement the function using C++ and show the source code listing.
b) We are also asked to add a function Position highestKey(const Position& v) const to the class
SearchTree.
Function prototype: Position highestKey(const Position& v) const;
input argument: position of the node in the binary search tree to calculate from. For the whole
tree this would be the position for the root of the tree.
return value: position of the node having the highest key value.
i. Describe strategy with reasonings and examples to convince that it will work.
ii. Implement the function using C++ and show the source code listing.
Solution
#include
#include
/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct node
{
int data;
struct node* left;
struct node* right;
};
/* Helper function that allocates a new node
with the given data and NULL left and right
pointers. */
struct node* newNode(int data)
{
struct node* node = (struct node*)
malloc(sizeof(struct node));
node->data = data;
node->left = NULL;
node->right = NULL;
return(node);
}
/* Give a binary search tree and a number,
inserts a new node with the given number in
the correct place in the tree. Returns the new
root pointer which the caller should then use
(the standard trick to avoid using reference
parameters). */
struct node* insert(struct node* node, int data)
{
/* 1. If the tree is empty, return a new,
single node */
if (node == NULL)
return(newNode(data));
else
{
/* 2. Otherwise, recur down the tree */
if (data <= node->data)
node->left = insert(node->left, data);
else
node->right = insert(node->right, data);
/* return the (unchanged) node pointer */
return node;
}
}
/* Given a non-empty binary search tree,
return the minimum data value found in that
tree. Note that the entire tree does not need
to be searched. */
int minValue(struct node* node) {
struct node* current = node;
/* loop down to find the leftmost leaf */
while (current->left != NULL) {
current = current->left;
}
return(current->data);
}
/* Driver program to test sameTree function*/
int main()
{
struct node* root = NULL;
root = insert(root, 4);
insert(root, 2);
insert(root, 1);
insert(root, 3);
insert(root, 6);
insert(root, 5);
printf(\"\ Minimum value in BST is %d\", minValue(root));
getchar();
return 0;
}
#include
#include
/* A binary tree node has data, pointer to left child
.
You can list anything, it doesnt matter. I just want to see code f.pdf
You can list anything, it doesn\'t matter. I just want to see code for this problem. Thank you
Solution
Include header files .......
Source code:
public class SinglyLinkedListImpl {
private Node head;
private Node tail;
public void add(T element){
Node nd = new Node();
nd.setValue(element);
System.out.println(\"Adding: \"+element);
if(head == null){
head = nd;
tail = nd;
} else {
tail.setNextRef(nd);
}
tail = nd;
}
}
public void addAfter(T element, T after){
Node tmp = head;
Node refNode = null;
System.out.println(\"Traversing to all nodes..\");
while(true){
if(tmp == null){
break;
}
if(tmp.compareTo(after) == 0){
refNode = tmp;
break;
}
tmp = tmp.getNextRef();
}
if(refNode != null){
Node nd = new Node();
nd.setValue(element);
nd.setNextRef(tmp.getNextRef());
if(tmp == tail){
tail = nd;
}
tmp.setNextRef(nd);
} else {
System.out.println(\"Unable to find the given element...\");
}
}
public void deleteFront(){
if(head == null){
System.out.println(\"Underflow...\");
}
Node tmp = head;
head = tmp.getNextRef();
if(head == null){
tail = null;
}
System.out.println(\"Deleted: \"+tmp.getValue());
}
public void deleteAfter(T after){
Node tmp = head;
Node refNode = null;
System.out.println(\"Traversing to all nodes..\");
while(true){
if(tmp == null){
break;
}
if(tmp.compareTo(after) == 0){
refNode = tmp;
break;
}
tmp = tmp.getNextRef();
}
if(refNode != null){
tmp = refNode.getNextRef();
refNode.setNextRef(tmp.getNextRef());
if(refNode.getNextRef() == null){
tail = refNode;
}
System.out.println(\"Deleted: \"+tmp.getValue());
} else {
System.out.println(\"Unable to find the given element...\");
}
}
public void traverse(){
Node tmp = head;
while(true){
if(tmp == null){
break;
}
System.out.println(tmp.getValue());
tmp = tmp.getNextRef();
}
}
public static void main(String a[]){
SinglyLinkedListImpl sl = new SinglyLinkedListImpl();
sl.add(3);
sl.add(32);
sl.add(54);
sl.add(89);
sl.addAfter(76, 54);
sl.deleteFront();
sl.deleteAfter(76);
sl.traverse();
}
}
class Node implements Comparable {
private T value;
private Node nextRef;
public T getValue() {
return value;
}
public void setValue(T value) {
this.value = value;
}
public Node getNextRef() {
return nextRef;
}
public void setNextRef(Node ref) {
this.nextRef = ref;
}
public int compareTo(T arg) {
if(arg == this.value){
return 0;
} else {
return 1;
}
}
}
This program implement the concept of single linked list..
Write a program that displays an AVL tree along with its balance fact.docx
Write a program that displays an AVL tree along with its balance fact
I have a .java program that I need to modify so that it1) reads i.pdf
I have a .java program that I need to modify so that it:
1) reads in a file called books.txt (i will provide a few lines of the text file below)
2) outputs a statement every time there's a collision (see the example output below in the original
prompt)
THIS IS THE ORIGINAL PROMPT:
Part A: Modify the attached code to build and customize a AVL tree of Book nodes. Using an
input file, insert Book nodes and detect imbalance. If imbalance is true, then call the proper
rotation function as discussed in the lecture to fix the imbalance condition.
Must read AVLNode data from a text file
Create a Book Object; and an AVL node object to be inserted into the AVL tree
At each insert, detect imbalance and fix the AVL tree
Report each imbalance detection and the node where it occurred; and output the message:
Example Output:
Imbalance condition occurred at inserting ISBN 12345; fixed in LeftRight Rotation
Imbalance condition occurred at inserting ISBN 87654; fixed in Left Rotation
Imbalance condition occurred at inserting ISBN 974321; fixed in RightLeft Rotation
class AVLNode {
Book bookObject;
int height;
AVLNode leftPtr;
AVLNode rightPtr;
}
BELOW IS THE CODE I NEED TO MODIFY:
// AVL Binary search tree implementation in Java
// Author: AlgorithmTutor
// data structure that represents a node in the tree
class Node {
int data; // holds the key
Node parent; // pointer to the parent
Node left; // pointer to left child
Node right; // pointer to right child
int bf; // balance factor of the node
public Node(int data) {
this.data = data;
this.parent = null;
this.left = null;
this.right = null;
this.bf = 0;
}
}
public class AVLTree {
private Node root;
public AVLTree() {
root = null;
}
private void printHelper(Node currPtr, String indent, boolean last) {
// print the tree structure on the screen
if (currPtr != null) {
System.out.print(indent);
if (last) {
System.out.print("R----");
indent += " ";
} else {
System.out.print("L----");
indent += "| ";
}
System.out.println(currPtr.data + "(BF = " + currPtr.bf + ")");
printHelper(currPtr.left, indent, false);
printHelper(currPtr.right, indent, true);
}
}
private Node searchTreeHelper(Node node, int key) {
if (node == null || key == node.data) {
return node;
}
if (key < node.data) {
return searchTreeHelper(node.left, key);
}
return searchTreeHelper(node.right, key);
}
private Node deleteNodeHelper(Node node, int key) {
// search the key
if (node == null) return node;
else if (key < node.data) node.left = deleteNodeHelper(node.left, key);
else if (key > node.data) node.right = deleteNodeHelper(node.right, key);
else {
// the key has been found, now delete it
// case 1: node is a leaf node
if (node.left == null && node.right == null) {
node = null;
}
// case 2: node has only one child
else if (node.left == null) {
Node temp = node;
node = node.right;
}
else if (node.right == null) {
Node temp = node;
node = node.left;
}
// case 3: has both children
else {
Node temp = minimum(node.right);
node.data = temp.data;
node.right = .
How to do the main method for this programBinaryNode.javapublic.pdf
How to do the main method for this program?
BinaryNode.java
public class BinaryNode
implements Parent{
/** The item associated with this node. */
private E item;
/** The node at the root of the left subtree. */
private BinaryNode left;
/** The node at the root of the right subtree. */
private BinaryNode right;
/** Put item in a leaf node. */
public BinaryNode(E item) {
this.item = item;
// left and right are set to null by default
}
/** no-argument constructor sets everything to null */
public BinaryNode() {
item = null;
left = null;
right = null;
}
/** Put item in a node with the specified subtrees. */
public BinaryNode(E item, BinaryNode left,
BinaryNode right) {
this.item = item;
this.left = left;
this.right = right;
}
/** Put item in a node with the specified subtrees. */
public BinaryNode getChild(int direction) {
if (direction < 0) {
return left;
} else {
return right;
}
}
/** Return the item associated with this node. */
public E getItem() {
return item;
}
/** Return the root of the left subtree. */
public BinaryNode getLeft() {
return left;
}
/** Return the root of the right subtree. */
public BinaryNode getRight() {
return right;
}
/** Return true if this is a leaf. */
public boolean isLeaf() {
return (left == null) && (right == null);
}
/** Replace the item associated with this node. */
public void setItem(E item) {
this.item = item;
}
/** Replace the left subtree with the one rooted at left. */
public void setLeft(BinaryNode left) {
this.left = left;
}
/** Replace the right subtree with the one rooted at right. */
public void setRight(BinaryNode right) {
this.right = right;
}
public void setChild(int direction, BinaryNode child) {
if (direction < 0) {
left = child;
} else {
right = child;
}
}
/**
* Return the String representation of the tree rooted at this node
* traversed preorder.
**/
public String toStringPreorder() {
String result = \"\";
result += item;
if (left != null) {
result += left.toStringPreorder();
}
if (right != null) {
result += right.toStringPreorder();
}
return result;
}
/** Return a String representation of the tree rooted at this node
* traversed inorder.
**/
public String toStringInOrder() {
String result = \"\";
if (left != null) {
result += left.toStringInOrder();
}
result += item;
if (right != null) {
result += right.toStringInOrder();
}
return result;
}
/** Return a String representation of the tree rooted at this node,
* traversed postorder.
**/
public String toStringPostorder() {
String result = \"\";
if (left != null) {
result += left.toStringPostorder();
}
if (right != null) {
result += right.toStringPostorder();
}
result += item;
return result;
}
/**
* Return a String representation of the tree rooted at this node,
* traversed level order.
**/
public String toStringLevelOrder() {
String result = \"\";
Queue> q = new ArrayQueue>();
q.add (this);
while (!(q.isEmpty())) {
BinaryNode node = q.remove ();
result += node.item;
if (node.left != null) {
q.add(node.left);
}
if (node.right != null) {
q.add(node..
public class Deque {private class Node {public int data;public.pdf
Given interval is `[0,2pi]` .
And given identity is `secx+tanx=1`
which we can write as `1/(cosx)+(sinx)/(cosx)=1`
`or` , `(1+sinx)/(cosx)=1`
or, `(1+sinx)=cosx`
or, `1+sinx-cosx=0`
or, `1+2sin(x/2)cos(x/2)-(1-2sin^2(x/2))=0`
or, `1+2sin(x/2)cos(x/2)-1+2sin^2(x/2)=0`
or, `2sin(x/2)cos(x/2)+2sin^2(x/2)=0`
or, `2sin(x/2)[cos(x/2)+sin(x/2)]=0`
`or,` `2sin(x/2)=0` and `cos(x/2)+sin(x/2)=0` .
Now, `2sin(x/2)=0` `rArrsin(x/2)=0` `rArrsin(x/2)=sin(npi)` , n=0,1,2. As per the need of our
problem.
`rArr x/2=npi` or, `x=2npi` for n=0,1 as our interval is `[0,2pi]` .
So, `x=0` or, `x=2pi` .
Now, if `cos(x/2)+sin(x/2)=0` `rArr1+(sin(x/2))/cos(x/2)=0`
`rArrtan(x/2)=-1` or, `tan(x/2)=tan(pi/2+pi/4)`
or, `x/2=(pi/2+pi/4)`
or, `x=pi+pi/2`
or, `x=3pi/2` .
Clearly, `x=0` and `x=2pi` are the solutions satisfying our given identity over the given interval.
Solution
Given interval is `[0,2pi]` .
And given identity is `secx+tanx=1`
which we can write as `1/(cosx)+(sinx)/(cosx)=1`
`or` , `(1+sinx)/(cosx)=1`
or, `(1+sinx)=cosx`
or, `1+sinx-cosx=0`
or, `1+2sin(x/2)cos(x/2)-(1-2sin^2(x/2))=0`
or, `1+2sin(x/2)cos(x/2)-1+2sin^2(x/2)=0`
or, `2sin(x/2)cos(x/2)+2sin^2(x/2)=0`
or, `2sin(x/2)[cos(x/2)+sin(x/2)]=0`
`or,` `2sin(x/2)=0` and `cos(x/2)+sin(x/2)=0` .
Now, `2sin(x/2)=0` `rArrsin(x/2)=0` `rArrsin(x/2)=sin(npi)` , n=0,1,2. As per the need of our
problem.
`rArr x/2=npi` or, `x=2npi` for n=0,1 as our interval is `[0,2pi]` .
So, `x=0` or, `x=2pi` .
Now, if `cos(x/2)+sin(x/2)=0` `rArr1+(sin(x/2))/cos(x/2)=0`
`rArrtan(x/2)=-1` or, `tan(x/2)=tan(pi/2+pi/4)`
or, `x/2=(pi/2+pi/4)`
or, `x=pi+pi/2`
or, `x=3pi/2` .
Clearly, `x=0` and `x=2pi` are the solutions satisfying our given identity over the given interval..
Create your own Linked list from scratch (singly linked list).So.pdf
Consider four protons at rest at the corners of a square. The protons are bound together so they
cannot move. Then the bonds holding one of the protons break so that proton is free to move.
Determine that proton\'s acceleration (both direction and magnitude) at the instant after it is no
longer bound as well as its speed when it has moved far away. You know the proton\'s charge, its
mass, and the length of the sides of the square.
Solution
Here we need to note that the like charges repel each other, hence the proton will be pushed
away along the diagonal of the square.
We will assume that the mass of the proton is Mp and charge is q while the length of the side of
the square is a.
The net force due to the proton diagonally opposite would be along the diagonal, and also due to
the other two charges.
Hence the net force would be: 2kq22/a2 + kq2/2a2 = (kq2/a2)[22 + 1/2] along the diagonal.
The acceleration would be given as: Force / mass = (kq2/Mpa2)[22 + 1/2]
So as to determine the speed of the proton once it gets far away, we will use the prinicple of
conservation of energy. Hence the initial potential energy of the proton would be equal to its
final kinetic energy hence giving us the speed.
That is. 2kq2/a + kq2/a2 = (kq2/a)[2 + 1/2] = Mpv2/2
or, V = sqrt[(2kq2/Mpa)[2 + 1/2]] which is the required expression for the speed of the proton..
In this lab, you will be given a simple code for a min Heap, and you.pdf
In this lab, you will be given a simple code for a min Heap, and youre going to complete some of
the methods to handle some edge cases. Tasks for today: There are mainly two tasks, and the
logic of both of them is provided in their bodies: 1) Fully implement AddItem(). 2) Fully
implement HeapUp() Use the following start-up code. The main code and a demo class are
provided as .cs files. Use them to test your code. Once you complete the above two tasks, upload
the source code for your MinHeap class to Blackboard. class BinaryHeap : IEnumerable where T
: IComparable { private T[] array; private int count; // Number of elements (nodes) public
BinaryHeap(int size) { array = new T[size]; count = 0; } public T GetItem(int index) { return
array[index]; } private void SetItem(int index, T value) { while (index >= array.Length)
Grow(array.Length * 2); array[index] = value; } private void Grow(int newsize) {
Array.Resize(ref array, newsize); } // Indices of left and right children private int
LeftChildIndex(int pos) { return 2 * pos + 1; } private int RightChildIndex(int pos) { return 2 *
pos + 2; } private int GetParentIndex(int pos) => (pos - 1) / 2; private T GetRightChild(int pos)
=> array[RightChildIndex(pos)]; private T GetLeftChild(int pos) => array[LeftChildIndex(pos)];
private T GetParent(int pos) => array[GetParentIndex(pos)]; // "Has" methods to determine if the
indices exist private bool HasLeftChild(int pos) { if (LeftChildIndex(pos) < count) return true;
else return false; } private bool HasRightChild(int pos) { if (RightChildIndex(pos) < count)
return true; else return false; } private bool IsRoot(int pos) => pos == 0; // (true if element is
root) // Swap, uh, swaps two values given two indicies. This should be private but I originally
had it public for some reason. private void Swap(int position1, int position2) { T first =
array[position1]; array[position1] = array[position2]; array[position2] = first; } public void
AddItem(T value) { // This is part of la //Insert Logic // If the tree is empty, insert at the bottom
(it does that already) // if not, insert at the end, // From the end you either need to swap to the
root, and keep minheapify (Heapdown) // or, you should probably implement move up (for lab 6,
you need to implement // for that you run a while loop, check if the current position both isn't the
root, and is higher priority than a parent (in this case that probably means it's a lower value) // if
it is, swap with parent, and keep doing that until it's its in the array[count] = value; count++;
ReCalculateUp(); } private void HeapDown() { //CompareTo //this.CompareTo(value) returns <
0 if this < value //this.CompareTo(value) returns >0 if this > value int index = 0; while
(HasLeftChild(index)) { var smallerIndex = LeftChildIndex(index); if (HasRightChild(index)
&& (GetRightChild(index).CompareTo(GetLeftChild(index))<0) ) //there's a set of ( ) around the
right expression that are redundant but hopefully easier to read { smallerI.
hi i have to write a java program involving link lists. i have a pro.pdf
hi i have to write a java program involving link lists. i have a problem with the nodes, as it is
posting errors with the nodes.
please help me with this problem. thank you.
public class LinkLists {
/* only need to store a single pointer to the node at the head
* of the list.
* The pointer is null if the list is empty.
* Also record the size of the list.
*/
protected Node head;
/* invariant: size is the number of nodes in the list pointed to by head */
protected int size;
/* no-arguments default constructor creates an empty list */
public LinkLists() {
head = null; // start with an empty list
size = 0;
}
/* accessor method */
public int size() {
return size;
}
/* value to add to the end of the list
*/
public void add(T value) {
head = addAtEnd(head, value);
size++;
}
/* node of the list to which the value should be added
* value to add to the end of the list
*/
private Node addAtEnd(Node node, T value) {
if (node == null) { // special case
return new Node(value, null);
} else if (node.getNext() == null) { // other special case
node.setNext(new Node(value, null));
} else {
addAtEnd(node.getNext(), value);
}
return node;
}
/* iterative implementation of the same method
* value to add to the end of the list
*/
public void add2(T value) {
if (head == null) {
head = new Node(value, null);
} else {
Node node = head; // guaranteed not to be null initially
while (node.getNext() != null) {
node = node.getNext(); // guaranteed not to be null here
}
// now, node.getNext() is guaranteed to be null
// similar to the second special case in addAtEnd
node.setNext(new Node(value, null));
}
size++;
}
public void remove(int position) throws BadItemCountException {
if ((position < 1) || (position > size)) {
throw new
BadItemCountException(\"invalid position \" + position +
\", only 1..\" + size + \" available\");
}
if (position == 1) {
head = head.getNext();
} else {
Node node = head;
for (int i = 2; i < position; i++) {
node = node.getNext();
}
node.setNext(node.getNext().getNext());
}
size--; // one less item
}
public String toString() {
return toString(head);
}
private String toString(Node node) {
if (node == null) {
return \"\";
} else {
return node.getValue() + \"\ \" + toString(node.getNext());
}
}
public static void main(String[] args) {
/* create two empty lists, make sure they print out correctly */
LinkLists list1 = new LinkLists();
LinkLists list2 = new LinkLists();
System.out.println(\"list1 = \'\" + list1 + \"\', list2 = \'\" + list2 + \"\'\");
System.out.println(\"list1.size() = \" + list1.size() +
\", list2.size() = \" + list2.size());
/* insert some items, keep checking */
list1.add(\"hello\");
list1.add(\"world\");
list2.add(\"foo\");
list2.add(\"bar\");
list2.add(\"baz\");
System.out.println(\"list1 = \'\" + list1 + \"\', list2 = \'\" + list2 + \"\'\");
System.out.println(\"list1.size() = \" + list1.size() +
\", list2.size() = \" + list2.size());
/* remove an item at an invalid position */
boolean caught = false;
try {
list2.remove(4);
} catch .
#include iostream using namespace std; const int nil = 0; cl.docx
#include iostream using namespace std; const int nil = 0; cl
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Given the following codepackage data1;import java.util.;p.pdf
Given the following code
package data1;
import java.util.*;
/*public class BST> implementing a set ADT and containing the following:
* private inner class BinaryNode representing a node with a (possibly) left child and a (possibly)
right child
** instance fields BinaryNode root, int size
* contains/insert/remove method - w/ O(height) complexity
** size O(1), isEmpty O(1), clear O(1),
*** findMin, findMax, findHeight methods - w/ O(height) complexity
**** toString() method printing set (in-order traversal) and the tree (BFS traversal)
**
*/
publicclass BST> {
private BinaryNode root;
privateint size;
//Search opertion of Set ADT
publicboolean contains(E value) {
return contains(root, value);
}
//Insert opertion of Set ADT
publicvoid insert(E value) {
root = insert(root, value);
size++;
}
//Remove operation of Set ADT
publicvoid remove(E value){
root = remove(root, value);
size--;
}
publicint size(){
return size;
}
publicboolean isEmpty(){
return root == null;
}
publicvoid clear(){
root = null;
size = 0;
}
public E findMin(){
return findMin(root);
}
public E findMax(){
return findMax(root);
}
publicint findHeight(){
return root.height();
}
public String toString(){
if(root == null)
return "set {} stored in an empty tree.\n";
String elements = root.inOrder().toString();
return "set {" + elements.substring(1,elements.length()-1) + "} stored in tree \n" + root;
}
private E findMin(BinaryNode node){
if(node == null)
returnnull;
if(node.left == null)
return node.element;
return findMin(node.left);
}
private E findMax(BinaryNode node){
if(node == null)
returnnull;
if(node.right == null)
return node.element;
return findMax(node.right);
}
privateboolean contains(BinaryNode node, E value) {
if (node == null)//node represents an empty substree
returnfalse;
int comparisonResult = value.compareTo(node.element);
if(comparisonResult < 0)//if value is less than root's value
return contains(node.left, value);
if(comparisonResult > 0)//if value is greater than root's value
return contains(node.right, value);
returntrue;//successful search
}
private BinaryNode insert(BinaryNode node, E value) {
if (node == null)//base case: empty tree
returnnew BinaryNode<>(value);
if (value.compareTo(node.element) < 0)
node.left = insert(node.left, value);//insert recursively to left-subtree
elseif (value.compareTo(node.element) > 0)
node.right = insert(node.right, value);//insert recursively to right-subtree
else {//duplicate value cannot be inserted in a BST implementing the Set ADT
size--;//insertion failed!
//System.out.print("BST.insert: Warning: " + value + " is already stored in the tree \n" + node);
}
return node;
}
private BinaryNode remove(BinaryNode node, E value) {
if (node == null) {//base case: empty tree
System.out.println("BST.remove: Warning: " + value + " doesn't exist in the tree.");
size++;//removal failed!
return node;
}
if (value.compareTo(node.element) < 0)
node.left = remove(node.left, value);
elseif (value.compareTo(node.element) > 0)
node.right = remov.
7- Match the astronomer to his accomplishment a- Ptolemy i- Made very.pdf
7. In humans, the allele for normal blood clotting, H , is dominant to the allele for hemophilia, h .
This is a sex-linked trait found on the X chromosome. A woman with normal blood clotting has
four children: a normal son, a hemophiliac son, and two normal daughters. The father has normal
blood clotting. What is the genotype for each member of the family? 8. Red-green color
blindness (b) is a recessive sex-linked trait. A colorblind male marries a normal female. Their
daughter is colorblind. a)What are the genotypes of both parents and the daughter? b) Can a
colorblind father have daughters who are not colorblind? Explain. c) If a normal sighted woman
whose father was color-blind marries a color-blind man, what is the probability that they will
have a colorblind child?.
7- In a multiple linear regression model y-X+-XN(0-2In)- the OLS estim (1).pdf
7. For the circuit in Figure 6-71, assume the inputs are A dd / S u b t . = 1 , A = 1001 and B =
1100 . What is the output?.
7- In addition to a pH imbalance- prolonged vomiting is also causing D.pdf
7. Identify multi-valued or composite fields, partial dependencies, transitive dependencies and
the current normal form for the following relational tables. a. BOOK (isbn title nublisher
address) Note: Book is published with a specific ISBN. The publisher can chanse addiesses after
book is published. c. BOOK (isbn, title, publisher) PUBLISHER (publisher, street1, street2, city,
state, zip) ivote: Ins example allows each publisher to have a single address. How would you
modify the table(s) if a publisher could have multiple addresses? d. ORDFR (nrder num nradunt
id nend Aacorintion) 8. Your customer had a spreadsheet with the following columns and sample
data. a. The person before you, who was fired, had created the following relational table in which
to import the spreadsheet. b. Explain the disadvantages of modeling the spreadsheet in this
manner. What are you going to do, if anything, to change this? CSCI 2020: Database
Management Page 16.
7- The temperature at which a certain substance boils is normally dist.pdf
7. The length of time necessary to complete a specific task is exponentially distributed with mean
= 0.5 hrs . If the task is to be done twice and the time to do the tasks are independent, what is the
probability that the total time for these two tasks will exceed 1.5 hours?.
7- The length of time necessary to complete a specific task is exponen.pdf
7. The diagrams below show two nonhomologous chromosomes. Each contains a copy of a
transposable element (orange). In panel A, the transposable element is inserted in the two
chromosomes in the same orientation. In panel B, the transposable element is oriented in
opposite directions. For both panels, draw diagrams illustrating the consequences of
recombination following alignment to these transposable elements (instead of homologous
chromosomes). 8. Reverse transcription is a phenomenon that is important in cellular processes
and biotechnology. How does it differ from transcription in gene expression? Give three
examples of reverse transcription. One as it is used in biotechnology. One as it operates enables
retrotransposons to fill the genome. One regarding its intergral role in the protection of linear
chromosomes..
7- The following data have been collected for a British healthcare IT.pdf
7. Suppose X 1 and X 2 are independent N ( , 1 ) random variables, for an unknown parameter .
(a) Find a sufficient statistic for . (b) Show that X 1 2 1 is an unbiased estimator of 2 . (c) Find a
MVUE estimator of 2 . You may use the fact that X 1 ( X 1 + X 2 = y ) N ( 2 y , 2 1 ) ..
7- Many news organizations conduct polls asking adults in the United S.pdf
7. In his book, Alfred Wegener presented many lines of evidence that Earth's continents have
"drifted" through time. Name and briefly explain four lines of evidence cited by Wegener. (6)
Since Wegener's death, many more types of data confirmed his ideas that continents are not
stationary. Names two important lines of evidence discovered/developed after Wegener's death.
(4).
7- Calculate - DV for the following foed label (For carbohydrate and F.pdf
7. Calculate % DV for the following foed label (For carbohydrate and Fat)..
7- With the help of suitable examples- Explain how the application of.pdf
7. Using the income elasticity of demand to characterize goods A survey taken by residents from
the imaginary town of Draw City tells economists that the following changes result from a 10%
fall in income: Compute the income elastioty of demand for each good and use the dropdown
menus to complete the first column in the following cable. Then, Bused on the income
elastioties, classily each good as either a normal good or an inferior good. (Hint: Be careful to
keep track of the direction of change. The sign of the income elssticity of demand can be postive
or negative, and the sign gives important information.) ework (Ch 05) - A 38% decrease in the
quantity of clubs demanded Compute the income elasticity of demand for each good and use the
dropdown menus to complete the first column in the following table. Then, based on the income
elasticties, classily each good as either a normal good or an infenor good. (Hint: Be careful to
keep track of the direction of change. The sign of the income elasticty of demand can be positive
or negative, and the sign gives important information.) Which of the following three goods is
most likely to be classified as a luxury good? Tokens Clabs Krigs.
7- A taxpayer's wife died in 2021 and he has not remarried- He has two.pdf
7. A taxpayer's wife died in 2021 and he has not remarried. He has two children, ages 6 and 8 ,
living at home. He provides all the support for his children. What is his most advantageous filing
status for 2022, 2023 and 2024 ? surviving spouse, surviving spouse, head of household
surviving spouse for all three years surviving spouse, surviving spouse, single head of household
for all three years.
7- In his book- Alfred Wegener presented many lines of evidence that E.pdf
7. If a lactase persistent person mated with a lactose intolerant person, what phenotype do you
think the off spring would have? Explain your reasoning. 8. Both A and B below refer to lactase.
Explain these figures and how they were derived. (A). (B).
More Related Content
Similar to Hello- the following code- In LeftRotaate and RightRotate methods for.pdf
Write a program that displays an AVL tree along with its balance fact.docx
Write a program that displays an AVL tree along with its balance fact
I have a .java program that I need to modify so that it1) reads i.pdf
I have a .java program that I need to modify so that it:
1) reads in a file called books.txt (i will provide a few lines of the text file below)
2) outputs a statement every time there's a collision (see the example output below in the original
prompt)
THIS IS THE ORIGINAL PROMPT:
Part A: Modify the attached code to build and customize a AVL tree of Book nodes. Using an
input file, insert Book nodes and detect imbalance. If imbalance is true, then call the proper
rotation function as discussed in the lecture to fix the imbalance condition.
Must read AVLNode data from a text file
Create a Book Object; and an AVL node object to be inserted into the AVL tree
At each insert, detect imbalance and fix the AVL tree
Report each imbalance detection and the node where it occurred; and output the message:
Example Output:
Imbalance condition occurred at inserting ISBN 12345; fixed in LeftRight Rotation
Imbalance condition occurred at inserting ISBN 87654; fixed in Left Rotation
Imbalance condition occurred at inserting ISBN 974321; fixed in RightLeft Rotation
class AVLNode {
Book bookObject;
int height;
AVLNode leftPtr;
AVLNode rightPtr;
}
BELOW IS THE CODE I NEED TO MODIFY:
// AVL Binary search tree implementation in Java
// Author: AlgorithmTutor
// data structure that represents a node in the tree
class Node {
int data; // holds the key
Node parent; // pointer to the parent
Node left; // pointer to left child
Node right; // pointer to right child
int bf; // balance factor of the node
public Node(int data) {
this.data = data;
this.parent = null;
this.left = null;
this.right = null;
this.bf = 0;
}
}
public class AVLTree {
private Node root;
public AVLTree() {
root = null;
}
private void printHelper(Node currPtr, String indent, boolean last) {
// print the tree structure on the screen
if (currPtr != null) {
System.out.print(indent);
if (last) {
System.out.print("R----");
indent += " ";
} else {
System.out.print("L----");
indent += "| ";
}
System.out.println(currPtr.data + "(BF = " + currPtr.bf + ")");
printHelper(currPtr.left, indent, false);
printHelper(currPtr.right, indent, true);
}
}
private Node searchTreeHelper(Node node, int key) {
if (node == null || key == node.data) {
return node;
}
if (key < node.data) {
return searchTreeHelper(node.left, key);
}
return searchTreeHelper(node.right, key);
}
private Node deleteNodeHelper(Node node, int key) {
// search the key
if (node == null) return node;
else if (key < node.data) node.left = deleteNodeHelper(node.left, key);
else if (key > node.data) node.right = deleteNodeHelper(node.right, key);
else {
// the key has been found, now delete it
// case 1: node is a leaf node
if (node.left == null && node.right == null) {
node = null;
}
// case 2: node has only one child
else if (node.left == null) {
Node temp = node;
node = node.right;
}
else if (node.right == null) {
Node temp = node;
node = node.left;
}
// case 3: has both children
else {
Node temp = minimum(node.right);
node.data = temp.data;
node.right = .
How to do the main method for this programBinaryNode.javapublic.pdf
How to do the main method for this program?
BinaryNode.java
public class BinaryNode
implements Parent{
/** The item associated with this node. */
private E item;
/** The node at the root of the left subtree. */
private BinaryNode left;
/** The node at the root of the right subtree. */
private BinaryNode right;
/** Put item in a leaf node. */
public BinaryNode(E item) {
this.item = item;
// left and right are set to null by default
}
/** no-argument constructor sets everything to null */
public BinaryNode() {
item = null;
left = null;
right = null;
}
/** Put item in a node with the specified subtrees. */
public BinaryNode(E item, BinaryNode left,
BinaryNode right) {
this.item = item;
this.left = left;
this.right = right;
}
/** Put item in a node with the specified subtrees. */
public BinaryNode getChild(int direction) {
if (direction < 0) {
return left;
} else {
return right;
}
}
/** Return the item associated with this node. */
public E getItem() {
return item;
}
/** Return the root of the left subtree. */
public BinaryNode getLeft() {
return left;
}
/** Return the root of the right subtree. */
public BinaryNode getRight() {
return right;
}
/** Return true if this is a leaf. */
public boolean isLeaf() {
return (left == null) && (right == null);
}
/** Replace the item associated with this node. */
public void setItem(E item) {
this.item = item;
}
/** Replace the left subtree with the one rooted at left. */
public void setLeft(BinaryNode left) {
this.left = left;
}
/** Replace the right subtree with the one rooted at right. */
public void setRight(BinaryNode right) {
this.right = right;
}
public void setChild(int direction, BinaryNode child) {
if (direction < 0) {
left = child;
} else {
right = child;
}
}
/**
* Return the String representation of the tree rooted at this node
* traversed preorder.
**/
public String toStringPreorder() {
String result = \"\";
result += item;
if (left != null) {
result += left.toStringPreorder();
}
if (right != null) {
result += right.toStringPreorder();
}
return result;
}
/** Return a String representation of the tree rooted at this node
* traversed inorder.
**/
public String toStringInOrder() {
String result = \"\";
if (left != null) {
result += left.toStringInOrder();
}
result += item;
if (right != null) {
result += right.toStringInOrder();
}
return result;
}
/** Return a String representation of the tree rooted at this node,
* traversed postorder.
**/
public String toStringPostorder() {
String result = \"\";
if (left != null) {
result += left.toStringPostorder();
}
if (right != null) {
result += right.toStringPostorder();
}
result += item;
return result;
}
/**
* Return a String representation of the tree rooted at this node,
* traversed level order.
**/
public String toStringLevelOrder() {
String result = \"\";
Queue> q = new ArrayQueue>();
q.add (this);
while (!(q.isEmpty())) {
BinaryNode node = q.remove ();
result += node.item;
if (node.left != null) {
q.add(node.left);
}
if (node.right != null) {
q.add(node..
public class Deque {private class Node {public int data;public.pdf
Given interval is `[0,2pi]` .
And given identity is `secx+tanx=1`
which we can write as `1/(cosx)+(sinx)/(cosx)=1`
`or` , `(1+sinx)/(cosx)=1`
or, `(1+sinx)=cosx`
or, `1+sinx-cosx=0`
or, `1+2sin(x/2)cos(x/2)-(1-2sin^2(x/2))=0`
or, `1+2sin(x/2)cos(x/2)-1+2sin^2(x/2)=0`
or, `2sin(x/2)cos(x/2)+2sin^2(x/2)=0`
or, `2sin(x/2)[cos(x/2)+sin(x/2)]=0`
`or,` `2sin(x/2)=0` and `cos(x/2)+sin(x/2)=0` .
Now, `2sin(x/2)=0` `rArrsin(x/2)=0` `rArrsin(x/2)=sin(npi)` , n=0,1,2. As per the need of our
problem.
`rArr x/2=npi` or, `x=2npi` for n=0,1 as our interval is `[0,2pi]` .
So, `x=0` or, `x=2pi` .
Now, if `cos(x/2)+sin(x/2)=0` `rArr1+(sin(x/2))/cos(x/2)=0`
`rArrtan(x/2)=-1` or, `tan(x/2)=tan(pi/2+pi/4)`
or, `x/2=(pi/2+pi/4)`
or, `x=pi+pi/2`
or, `x=3pi/2` .
Clearly, `x=0` and `x=2pi` are the solutions satisfying our given identity over the given interval.
Solution
Given interval is `[0,2pi]` .
And given identity is `secx+tanx=1`
which we can write as `1/(cosx)+(sinx)/(cosx)=1`
`or` , `(1+sinx)/(cosx)=1`
or, `(1+sinx)=cosx`
or, `1+sinx-cosx=0`
or, `1+2sin(x/2)cos(x/2)-(1-2sin^2(x/2))=0`
or, `1+2sin(x/2)cos(x/2)-1+2sin^2(x/2)=0`
or, `2sin(x/2)cos(x/2)+2sin^2(x/2)=0`
or, `2sin(x/2)[cos(x/2)+sin(x/2)]=0`
`or,` `2sin(x/2)=0` and `cos(x/2)+sin(x/2)=0` .
Now, `2sin(x/2)=0` `rArrsin(x/2)=0` `rArrsin(x/2)=sin(npi)` , n=0,1,2. As per the need of our
problem.
`rArr x/2=npi` or, `x=2npi` for n=0,1 as our interval is `[0,2pi]` .
So, `x=0` or, `x=2pi` .
Now, if `cos(x/2)+sin(x/2)=0` `rArr1+(sin(x/2))/cos(x/2)=0`
`rArrtan(x/2)=-1` or, `tan(x/2)=tan(pi/2+pi/4)`
or, `x/2=(pi/2+pi/4)`
or, `x=pi+pi/2`
or, `x=3pi/2` .
Clearly, `x=0` and `x=2pi` are the solutions satisfying our given identity over the given interval..
Create your own Linked list from scratch (singly linked list).So.pdf
Consider four protons at rest at the corners of a square. The protons are bound together so they
cannot move. Then the bonds holding one of the protons break so that proton is free to move.
Determine that proton\'s acceleration (both direction and magnitude) at the instant after it is no
longer bound as well as its speed when it has moved far away. You know the proton\'s charge, its
mass, and the length of the sides of the square.
Solution
Here we need to note that the like charges repel each other, hence the proton will be pushed
away along the diagonal of the square.
We will assume that the mass of the proton is Mp and charge is q while the length of the side of
the square is a.
The net force due to the proton diagonally opposite would be along the diagonal, and also due to
the other two charges.
Hence the net force would be: 2kq22/a2 + kq2/2a2 = (kq2/a2)[22 + 1/2] along the diagonal.
The acceleration would be given as: Force / mass = (kq2/Mpa2)[22 + 1/2]
So as to determine the speed of the proton once it gets far away, we will use the prinicple of
conservation of energy. Hence the initial potential energy of the proton would be equal to its
final kinetic energy hence giving us the speed.
That is. 2kq2/a + kq2/a2 = (kq2/a)[2 + 1/2] = Mpv2/2
or, V = sqrt[(2kq2/Mpa)[2 + 1/2]] which is the required expression for the speed of the proton..
In this lab, you will be given a simple code for a min Heap, and you.pdf
In this lab, you will be given a simple code for a min Heap, and youre going to complete some of
the methods to handle some edge cases. Tasks for today: There are mainly two tasks, and the
logic of both of them is provided in their bodies: 1) Fully implement AddItem(). 2) Fully
implement HeapUp() Use the following start-up code. The main code and a demo class are
provided as .cs files. Use them to test your code. Once you complete the above two tasks, upload
the source code for your MinHeap class to Blackboard. class BinaryHeap : IEnumerable where T
: IComparable { private T[] array; private int count; // Number of elements (nodes) public
BinaryHeap(int size) { array = new T[size]; count = 0; } public T GetItem(int index) { return
array[index]; } private void SetItem(int index, T value) { while (index >= array.Length)
Grow(array.Length * 2); array[index] = value; } private void Grow(int newsize) {
Array.Resize(ref array, newsize); } // Indices of left and right children private int
LeftChildIndex(int pos) { return 2 * pos + 1; } private int RightChildIndex(int pos) { return 2 *
pos + 2; } private int GetParentIndex(int pos) => (pos - 1) / 2; private T GetRightChild(int pos)
=> array[RightChildIndex(pos)]; private T GetLeftChild(int pos) => array[LeftChildIndex(pos)];
private T GetParent(int pos) => array[GetParentIndex(pos)]; // "Has" methods to determine if the
indices exist private bool HasLeftChild(int pos) { if (LeftChildIndex(pos) < count) return true;
else return false; } private bool HasRightChild(int pos) { if (RightChildIndex(pos) < count)
return true; else return false; } private bool IsRoot(int pos) => pos == 0; // (true if element is
root) // Swap, uh, swaps two values given two indicies. This should be private but I originally
had it public for some reason. private void Swap(int position1, int position2) { T first =
array[position1]; array[position1] = array[position2]; array[position2] = first; } public void
AddItem(T value) { // This is part of la //Insert Logic // If the tree is empty, insert at the bottom
(it does that already) // if not, insert at the end, // From the end you either need to swap to the
root, and keep minheapify (Heapdown) // or, you should probably implement move up (for lab 6,
you need to implement // for that you run a while loop, check if the current position both isn't the
root, and is higher priority than a parent (in this case that probably means it's a lower value) // if
it is, swap with parent, and keep doing that until it's its in the array[count] = value; count++;
ReCalculateUp(); } private void HeapDown() { //CompareTo //this.CompareTo(value) returns <
0 if this < value //this.CompareTo(value) returns >0 if this > value int index = 0; while
(HasLeftChild(index)) { var smallerIndex = LeftChildIndex(index); if (HasRightChild(index)
&& (GetRightChild(index).CompareTo(GetLeftChild(index))<0) ) //there's a set of ( ) around the
right expression that are redundant but hopefully easier to read { smallerI.
hi i have to write a java program involving link lists. i have a pro.pdf
hi i have to write a java program involving link lists. i have a problem with the nodes, as it is
posting errors with the nodes.
please help me with this problem. thank you.
public class LinkLists {
/* only need to store a single pointer to the node at the head
* of the list.
* The pointer is null if the list is empty.
* Also record the size of the list.
*/
protected Node head;
/* invariant: size is the number of nodes in the list pointed to by head */
protected int size;
/* no-arguments default constructor creates an empty list */
public LinkLists() {
head = null; // start with an empty list
size = 0;
}
/* accessor method */
public int size() {
return size;
}
/* value to add to the end of the list
*/
public void add(T value) {
head = addAtEnd(head, value);
size++;
}
/* node of the list to which the value should be added
* value to add to the end of the list
*/
private Node addAtEnd(Node node, T value) {
if (node == null) { // special case
return new Node(value, null);
} else if (node.getNext() == null) { // other special case
node.setNext(new Node(value, null));
} else {
addAtEnd(node.getNext(), value);
}
return node;
}
/* iterative implementation of the same method
* value to add to the end of the list
*/
public void add2(T value) {
if (head == null) {
head = new Node(value, null);
} else {
Node node = head; // guaranteed not to be null initially
while (node.getNext() != null) {
node = node.getNext(); // guaranteed not to be null here
}
// now, node.getNext() is guaranteed to be null
// similar to the second special case in addAtEnd
node.setNext(new Node(value, null));
}
size++;
}
public void remove(int position) throws BadItemCountException {
if ((position < 1) || (position > size)) {
throw new
BadItemCountException(\"invalid position \" + position +
\", only 1..\" + size + \" available\");
}
if (position == 1) {
head = head.getNext();
} else {
Node node = head;
for (int i = 2; i < position; i++) {
node = node.getNext();
}
node.setNext(node.getNext().getNext());
}
size--; // one less item
}
public String toString() {
return toString(head);
}
private String toString(Node node) {
if (node == null) {
return \"\";
} else {
return node.getValue() + \"\ \" + toString(node.getNext());
}
}
public static void main(String[] args) {
/* create two empty lists, make sure they print out correctly */
LinkLists list1 = new LinkLists();
LinkLists list2 = new LinkLists();
System.out.println(\"list1 = \'\" + list1 + \"\', list2 = \'\" + list2 + \"\'\");
System.out.println(\"list1.size() = \" + list1.size() +
\", list2.size() = \" + list2.size());
/* insert some items, keep checking */
list1.add(\"hello\");
list1.add(\"world\");
list2.add(\"foo\");
list2.add(\"bar\");
list2.add(\"baz\");
System.out.println(\"list1 = \'\" + list1 + \"\', list2 = \'\" + list2 + \"\'\");
System.out.println(\"list1.size() = \" + list1.size() +
\", list2.size() = \" + list2.size());
/* remove an item at an invalid position */
boolean caught = false;
try {
list2.remove(4);
} catch .
#include iostream using namespace std; const int nil = 0; cl.docx
#include iostream using namespace std; const int nil = 0; cl
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Given the following codepackage data1;import java.util.;p.pdf
Given the following code
package data1;
import java.util.*;
/*public class BST> implementing a set ADT and containing the following:
* private inner class BinaryNode representing a node with a (possibly) left child and a (possibly)
right child
** instance fields BinaryNode root, int size
* contains/insert/remove method - w/ O(height) complexity
** size O(1), isEmpty O(1), clear O(1),
*** findMin, findMax, findHeight methods - w/ O(height) complexity
**** toString() method printing set (in-order traversal) and the tree (BFS traversal)
**
*/
publicclass BST> {
private BinaryNode root;
privateint size;
//Search opertion of Set ADT
publicboolean contains(E value) {
return contains(root, value);
}
//Insert opertion of Set ADT
publicvoid insert(E value) {
root = insert(root, value);
size++;
}
//Remove operation of Set ADT
publicvoid remove(E value){
root = remove(root, value);
size--;
}
publicint size(){
return size;
}
publicboolean isEmpty(){
return root == null;
}
publicvoid clear(){
root = null;
size = 0;
}
public E findMin(){
return findMin(root);
}
public E findMax(){
return findMax(root);
}
publicint findHeight(){
return root.height();
}
public String toString(){
if(root == null)
return "set {} stored in an empty tree.\n";
String elements = root.inOrder().toString();
return "set {" + elements.substring(1,elements.length()-1) + "} stored in tree \n" + root;
}
private E findMin(BinaryNode node){
if(node == null)
returnnull;
if(node.left == null)
return node.element;
return findMin(node.left);
}
private E findMax(BinaryNode node){
if(node == null)
returnnull;
if(node.right == null)
return node.element;
return findMax(node.right);
}
privateboolean contains(BinaryNode node, E value) {
if (node == null)//node represents an empty substree
returnfalse;
int comparisonResult = value.compareTo(node.element);
if(comparisonResult < 0)//if value is less than root's value
return contains(node.left, value);
if(comparisonResult > 0)//if value is greater than root's value
return contains(node.right, value);
returntrue;//successful search
}
private BinaryNode insert(BinaryNode node, E value) {
if (node == null)//base case: empty tree
returnnew BinaryNode<>(value);
if (value.compareTo(node.element) < 0)
node.left = insert(node.left, value);//insert recursively to left-subtree
elseif (value.compareTo(node.element) > 0)
node.right = insert(node.right, value);//insert recursively to right-subtree
else {//duplicate value cannot be inserted in a BST implementing the Set ADT
size--;//insertion failed!
//System.out.print("BST.insert: Warning: " + value + " is already stored in the tree \n" + node);
}
return node;
}
private BinaryNode remove(BinaryNode node, E value) {
if (node == null) {//base case: empty tree
System.out.println("BST.remove: Warning: " + value + " doesn't exist in the tree.");
size++;//removal failed!
return node;
}
if (value.compareTo(node.element) < 0)
node.left = remove(node.left, value);
elseif (value.compareTo(node.element) > 0)
node.right = remov.
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Digital Tools and AI for Teaching Learning and Research
This Presentation in details discusses on Digital Tools and AI for Teaching Learning and Research
Chapter 3 - Islamic Banking Products and Services.pptx
Chapter 3 - Islamic Banking Products and Services.pptxMohd Adib Abd Muin, Senior Lecturer at Universiti Utara Malaysia
This slide is prepared for master's students (MIFB & MIBS) UUM. May it be useful to all.BÀI TẬP BỔ TRỢ TIẾNG ANH GLOBAL SUCCESS LỚP 3 - CẢ NĂM (CÓ FILE NGHE VÀ ĐÁP Á...
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https://app.box.com/s/hqnndn05v4q5a4k4jd597rkdbda0fniiFrancesca Gottschalk - How can education support child empowerment.pptx
Francesca Gottschalk from the OECD’s Centre for Educational Research and Innovation presents at the Ask an Expert Webinar: How can education support child empowerment?
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Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
Chapter -12, Antibiotics (One Page Notes).pdf
This is a notes for the D.Pharm students and related to the antibiotic drugs.
The French Revolution Class 9 Study Material pdf free download
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
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Child centred education is an educational approach that priorities the interest, needs and abilities of the child in the learning process.
"Protectable subject matters, Protection in biotechnology, Protection of othe...
Protectable subject matters, Protection in biotechnology, Protection of other biological materials, Ownership and period of protection
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Digital Tools and AI for Teaching Learning and Research
Digital Tools and AI for Teaching Learning and Research
Chapter 3 - Islamic Banking Products and Services.pptx
Chapter 3 - Islamic Banking Products and Services.pptx
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Francesca Gottschalk - How can education support child empowerment.pptx
Francesca Gottschalk - How can education support child empowerment.pptx
The French Revolution Class 9 Study Material pdf free download
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1.4 modern child centered education - mahatma gandhi-2.pptx
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Hello- the following code- In LeftRotaate and RightRotate methods for.pdf
- 1. Hello, the following code. In LeftRotaate and RightRotate methods for all occurrences of "parent", it errors indicating that "parent cannot be resolved or not a filed". Can you help me to correct this. public boolean insert(int i) { if(root == null) { root = new Node(i, Color.BLACK); return true; } Stack<Node> stack = new Stack<Node>(); Node node = root; while(node != NIL) { if(i == node.data) { return false; // duplicate value } stack.push(node); if(i < node.data) { node = node.left; } else { node = node.right; } } Node parent = stack.peek(); Node newNode = new Node(i, Color.RED); if(i < parent.data) {
- 2. parent.left = newNode; } else { parent.right = newNode; } insertFixup(newNode, stack); return true; } private void insertFixup(Node node, Stack<Node> stack) { Node parent, grandParent, uncle; while(node.color == Color.RED && (parent = stack.pop()).color == Color.RED) { grandParent = stack.empty() ? null : stack.peek(); if(grandParent == null) { node.color = Color.BLACK; } else { uncle = grandParent.left == parent ? grandParent.right : grandParent.left; if(uncle.color == Color.RED) { parent.color = uncle.color = Color.BLACK; grandParent.color = Color.RED; node = grandParent; } else { if(parent.right == node && grandParent.left == parent) { leftRotate(parent); node = parent;
- 3. parent = node.left; } else if(parent.left == node && grandParent.right == parent) { rightRotate(parent); node = parent; parent = node.right; } parent.color = Color.BLACK; grandParent.color = Color.RED; if(parent.left == node && grandParent.left == parent) { rightRotate(grandParent); } else { leftRotate(grandParent); } } } } root.color = Color.BLACK; } void leftRotate(Node node) { Node parent = node.right; node.right = parent.left; if(parent.left != NIL) { parent.left = node.right;
- 4. } if(node != root) { if(node == node.parent.left) { node.parent.left = parent; } else { node.parent.right = parent; } } else { root = parent; } parent.left = node; node.parent = parent; } private void rightRotate(Node node) { Node parent = node.left; node.left = parent.right; if(parent.right != NIL) { parent.right.parent = node; } if(node != root) { if(node == node.parent.right) { node.parent.right = parent; } else {
- 5. node.parent.left = parent; } } else { root = parent; } parent.right = node; node.parent = parent; }