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- 1. Elementary Data Structures & Applications By: Kuber Chandra Student B.tech 3rd yr(CSE) Babu Banarsi Das Institute, Ghaziabad Tuesday, April 8, 2014 1
- 2. Data Structure is the structural representation of logical relationships between elements of data. Data Structure = Organized Data + Operations Data Structure + Algorithm = Program Algorithm is a step-by-step finite sequence of instructions, to solve a well-defined computational problem. •Data Structure (In Memory) = Storage Structure •Data Structure (In Auxiliary Memory) = File Structure Complexity Analysis of Algorithm Time Complexity* Space Complexity** * Time depends on Algorithms, Languages, Compilers, CPU Speed etc. It is analyzed for best, average and worst case of input data. ** Space needed by instruction space, data space and environ. Stack space. Amstrong Complexity (amortized) Tuesday, April 8, 2014 2
- 3. Tuesday, April 8, 2014 3 Data Structure Primitive DS Non-primitive DS Integer Float Character Pointer Arrays Lists Files Linear List Non-linear List Stacks Queues Graphs Trees
- 4. Tuesday, April 8, 2014 4 ARRAYS An array is a collection of homogeneous data elements described by a single name. Each element is referenced by a subscripted variable or value, called subscript or index enclosed in parenthesis. •If an element is referenced by a single subscript, then array is known as 1-D or single array (linear array). --- Array[N] •If two subscripts are required, the array is known as 2-D or matrix array. ---- Array[M][N] •An array referenced by two or more subscripts is known as multidimensional array. ----- Array[X][Y][Z] Sparse array is an application of arrays where nearly all the elements have same values (usually zeros) and this value is constant. 1-D sparse array is called sparse vector and 2-D sparse array is called sparse matrix.
- 5. Tuesday, April 8, 2014 5 LISTS A list is a ordered set consisting of a varying number of elements (or nodes) linked by means of pointers. Each node is divided into two parts: A list overcome the limitation of fixed size in an array. A list may be linear ( stack, Queue) or non-linear (Tree, Graph). NODE Types of linked list (LL): 1. Singly LL 2. Doubly LL 3. Circular LL INFO LINK DATA LINK DATA LINK DATA LINK DATA LINK START
- 6. Tuesday, April 8, 2014 6 Types of linked lists 1. Singly linked list Begins with a pointer to the first node Terminates with a null pointer Only traversed in one direction 2. Circular, singly linked list (e.g., time sharing in OS, multiplayer game) Pointer in the last node points back to the first node 3. Doubly linked list (e.g., applications need more deletions) Two “start pointers” – first element and last element Each node has a forward pointer and a backward pointer Allows traversals both forwards and backwards 4. Circular, doubly linked list Forward pointer of the last node points to the first node and backward pointer of the first node points to the last node * Free storage, garbage collection, Dangling reference, reference counter, storage compaction
- 7. Tuesday, April 8, 2014 7 STACKS A stack is a non-primitive linear data structure (LIFO). It is an ordered collections of items where insertion and deletion take place at one end only called top of stack (TOS). Stack Operations: Push() Pop() Top() Size() IsEmpty() Stack Applications: •Page-visited history in a Web browser •Undo sequence in a text editor •Saving local variables when there is recursive function calls •Conversion of infix to postfix expression •Auxiliary data structure for algorithms •Component of other data structures
- 8. Tuesday, April 8, 2014 8 1 /* stack.c 2 dynamic stack program */ 3 #include <stdio.h> 4 #include <stdlib.h> 5 6 struct stackNode { /* self-referential structure */7 int data; 8 struct stackNode *nextPtr; 9 }; 10 11 typedef struct stackNode StackNode; 12 typedef StackNode *StackNodePtr; 13 14 void push( StackNodePtr *, int ); 15 int pop( StackNodePtr * ); 16 int isEmpty( StackNodePtr ); 17 void printStack( StackNodePtr ); 18 void instructions( void ); 19 20 int main() 21 { 22 StackNodePtr stackPtr = NULL; /* points to stack top */23 int choice, value; 24 25 instructions(); 26 printf( "? " ); 27 scanf( "%d", &choice ); 28 Outline 1. Define struct 1.1 Function definitions 1.2 Initialize variables 2. Input choice
- 9. Tuesday, April 8, 2014 9 29 while ( choice != 3 ) { 30 31 switch ( choice ) { 32 case 1: /* push value onto stack */ 33 printf( "Enter an integer: " ); 34 scanf( "%d", &value ); 35 push( &stackPtr, value ); 36 printStack( stackPtr ); 37 break; 38 case 2: /* pop value off stack */ 39 if ( !isEmpty( stackPtr ) ) 40 printf( "The popped value is %d.n", 41 pop( &stackPtr ) ); 42 43 printStack( stackPtr ); 44 break; 45 default: 46 printf( "Invalid choice.nn" ); 47 instructions(); 48 break; 49 } 50 51 printf( "? " ); 52 scanf( "%d", &choice ); 53 } 54 55 printf( "End of run.n" ); 56 return 0; 57 } 58 Outline 2.1 switch statement
- 10. Tuesday, April 8, 2014 10 59 /* Print the instructions */ 60 void instructions( void ) 61 { 62 printf( "Enter choice:n" 63 "1 to push a value on the stackn" 64 "2 to pop a value off the stackn" 65 "3 to end programn" ); 66 } 67 68 /* Insert a node at the stack top */ 69 void push( StackNodePtr *topPtr, int info ) 70 { 71 StackNodePtr newPtr; 72 73 newPtr = malloc( sizeof( StackNode ) ); 74 if ( newPtr != NULL ) { 75 newPtr->data = info; 76 newPtr->nextPtr = *topPtr; 77 *topPtr = newPtr; 78 } 79 else 80 printf( "%d not inserted. No memory available.n",81 info ); 82 } 83 Outline 3. Function definitions
- 11. Tuesday, April 8, 2014 11 84 /* Remove a node from the stack top */ 85 int pop( StackNodePtr *topPtr ) 86 { 87 StackNodePtr tempPtr; 88 int popValue; 89 90 tempPtr = *topPtr; 91 popValue = ( *topPtr )->data; 92 *topPtr = ( *topPtr )->nextPtr; 93 free( tempPtr ); 94 return popValue; 95 } 96 97 /* Print the stack */ 98 void printStack( StackNodePtr currentPtr ) 99 { 100 if ( currentPtr == NULL ) 101 printf( "The stack is empty.nn" ); 102 else { 103 printf( "The stack is:n" ); 104 105 while ( currentPtr != NULL ) { 106 printf( "%d --> ", currentPtr->data ); 107 currentPtr = currentPtr->nextPtr; 108 } 109 110 printf( "NULLnn" ); 111 } 112 } 113 Outline 3. Function definitions
- 12. Tuesday, April 8, 2014 12 114 /* Is the stack empty? */ 115 int isEmpty( StackNodePtr topPtr ) 116 { 117 return topPtr == NULL; 118 } Enter choice: 1 to push a value on the stack 2 to pop a value off the stack 3 to end program ? 1 Enter an integer: 5 The stack is: 5 --> NULL ? 1 Enter an integer: 6 The stack is: 6 --> 5 --> NULL ? 1 Enter an integer: 4 The stack is: 4 --> 6 --> 5 --> NULL ? 2 The popped value is 4. The stack is: 6 --> 5 --> NULL Outline 3. Function definitions Program Output
- 13. Tuesday, April 8, 2014 13 ? 2 The popped value is 6. The stack is: 5 --> NULL ? 2 The popped value is 5. The stack is empty. ? 2 The stack is empty. ? 4 Invalid choice. Enter choice: 1 to push a value on the stack 2 to pop a value off the stack 3 to end program ? 3 End of run. Outline Program Output
- 14. Tuesday, April 8, 2014 14 Tower of Hanoi: Application of stack History from Kashi Vishwanath temple which contains a large room with three time-worn posts in it surrounded by 64 golden disks. Brahmin priests, acting out the command of an ancient prophecy, have been moving these disks, in accordance with the immutable rules of the Brahma, since that time. The puzzle is therefore also known as the Tower of Brahma puzzle. According to the legend, when the last move of the puzzle will be completed, the world will end. If the legend were true, and if the priests were able to move disks at a rate of one per second, using the smallest number of moves, it would take them 264−1 seconds or roughly 585 billion years or 18,446,744,073,709,551,615 turns to finish, or about 45 times the life span of the sun.
- 15. Tuesday, April 8, 2014 15 Tower of Hanoi: Application of stack Three pegs A, B, C are given. Peg A contains N disks with decreasing size. The objective is to move disks from A to C using B. Step 1 Step 2 Step 3 Tower(N, A, B, C) If N=1 Tower(1, A, B, C) or A->C If N>1 1. Tower(N-1, A, C, B) 2. Tower(1, A, B, C) or A->C 3. Tower(N-1, B, A, C)
- 16. Tuesday, April 8, 2014 16 STACKS - Excercise Describe the output of the following series of stack operations •Push(8) •Push(3) •Pop() •Push(2) •Push(5) •Pop() •Pop() •Push(9) •Push(1) 3 8 5 2 8 1 9 8
- 17. Tuesday, April 8, 2014 17 QUEUES A queue is a non-primitive linear data structure (FIFO). It is an ordered collections of items where insertion takes place at one end called rear and deletion takes place at other end called front. Queue Operations: Enqueue() Dequeue() Front() Size() IsEmpty() Queue Applications: Direct applications •Waiting lines •Access to shared resources (e.g., printer) •Multiprogramming Indirect applications •Auxiliary data structure for algorithms •Component of other data structures
- 18. Tuesday, April 8, 2014 18 1 /* queue.c 2 Operating and maintaining a queue */ 3 4 #include <stdio.h> 5 #include <stdlib.h> 6 7 struct queueNode { /* self-referential structure */ 8 char data; 9 struct queueNode *nextPtr; 10 }; 11 12 typedef struct queueNode QueueNode; 13 typedef QueueNode *QueueNodePtr; 14 15 /* function prototypes */ 16 void printQueue( QueueNodePtr ); 17 int isEmpty( QueueNodePtr ); 18 char dequeue( QueueNodePtr *, QueueNodePtr * ); 19 void enqueue( QueueNodePtr *, QueueNodePtr *, char ); 20 void instructions( void ); 21 22 int main() 23 { 24 QueueNodePtr headPtr = NULL, tailPtr = NULL; 25 int choice; 26 char item; 27 28 instructions(); 29 printf( "? " ); 30 scanf( "%d", &choice ); Outline 1. Define struct 1.1 Function prototypes 1.2 Initialize variables 2. Input choice
- 19. Tuesday, April 8, 2014 19 31 32 while ( choice != 3 ) { 33 34 switch( choice ) { 35 36 case 1: 37 printf( "Enter a character: " ); 38 scanf( "n%c", &item ); 39 enqueue( &headPtr, &tailPtr, item ); 40 printQueue( headPtr ); 41 break; 42 case 2: 43 if ( !isEmpty( headPtr ) ) { 44 item = dequeue( &headPtr, &tailPtr ); 45 printf( "%c has been dequeued.n", item ); 46 } 47 48 printQueue( headPtr ); 49 break; 50 51 default: 52 printf( "Invalid choice.nn" ); 53 instructions(); 54 break; 55 } 56 57 printf( "? " ); 58 scanf( "%d", &choice ); 59 } 60 61 printf( "End of run.n" ); 62 return 0; 63 } 64 Outline 2.1 Switch statement
- 20. Tuesday, April 8, 2014 20 65 void instructions( void ) 66 { 67 printf ( "Enter your choice:n" 68 " 1 to add an item to the queuen" 69 " 2 to remove an item from the queuen" 70 " 3 to endn" ); 71 } 72 73 void enqueue( QueueNodePtr *headPtr, QueueNodePtr *tailPtr, 74 char value ) 75 { 76 QueueNodePtr newPtr; 77 78 newPtr = malloc( sizeof( QueueNode ) ); 79 80 if ( newPtr != NULL ) { 81 newPtr->data = value; 82 newPtr->nextPtr = NULL; 83 84 if ( isEmpty( *headPtr ) ) 85 *headPtr = newPtr; 86 else 87 ( *tailPtr )->nextPtr = newPtr; 88 89 *tailPtr = newPtr; 90 } 91 else 92 printf( "%c not inserted. No memory available.n", 93 value ); 94 } 95 Outline 3 function definitions
- 21. Tuesday, April 8, 2014 21 96 char dequeue( QueueNodePtr *headPtr, QueueNodePtr *tailPtr )97 { 98 char value; 99 QueueNodePtr tempPtr; 100 101 value = ( *headPtr )->data; 102 tempPtr = *headPtr; 103 *headPtr = ( *headPtr )->nextPtr; 104 105 if ( *headPtr == NULL ) 106 *tailPtr = NULL; 107 108 free( tempPtr ); 109 return value; 110 } 111 112 int isEmpty( QueueNodePtr headPtr ) 113 { 114 return headPtr == NULL; 115 } 116 117 void printQueue( QueueNodePtr currentPtr ) 118 { 119 if ( currentPtr == NULL ) 120 printf( "Queue is empty.nn" ); 121 else { 122 printf( "The queue is:n" ); Outline 3 function definitions
- 22. Tuesday, April 8, 2014 22 123 124 while ( currentPtr != NULL ) { 125 printf( "%c --> ", currentPtr->data ); 126 currentPtr = currentPtr->nextPtr; 127 } 128 129 printf( "NULLnn" ); 130 } 131 } Enter your choice: 1 to add an item to the queue 2 to remove an item from the queue 3 to end ? 1 Enter a character: A The queue is: A --> NULL ? 1 Enter a character: B The queue is: A --> B --> NULL ? 1 Enter a character: C The queue is: A --> B --> C --> NULL Outline 3 function definitions Output
- 23. Tuesday, April 8, 2014 23 ? 2 A has been dequeued. The queue is: B --> C --> NULL ? 2 B has been dequeued. The queue is: C --> NULL ? 2 C has been dequeued. Queue is empty. ? 2 Queue is empty. ? 4 Invalid choice. Enter your choice: 1 to add an item to the queue 2 to remove an item from the queue 3 to end ? 3 End of run. Outline Output
- 24. Tuesday, April 8, 2014 24 Types of Queues 1. Circular Queue Elements are represented in circular fashion. Insertion is done at very first location if last location is full. Rear=(rear + 1) % Size Front=(Front +1) % Size 2. Double Ended Queue (deque) Elements can be inserted or deleted from both ends. Input restricted deque-insertion at only one end Output restricted deque-deletion at only one end 3. Priority Queue Each element is assigned with a priority An element with higher priority is processed first Two elements with same priority are processed in FIFO order
- 25. Tuesday, April 8, 2014 254/8/2014 25 QUEUES - Excercise Describe the output of the following series of queue operations •enqueue(8) •enqueue(3) •dequeue() •enqueue(2) •enqueue(5) •dequeue() •dequeue() •enqueue(9) •enqueue(1) 8 3 3 2 5 5 9 1
- 26. Tuesday, April 8, 2014 26 The Trees A tree is a hierarchical representation of a finite set of one or more data items such that: •There is a special node called the root of the tree. •Data items are partitioned into mutually exclusive subsets each of which is itself a sub tree. Trees 2-way tree (binary tree) m-way tree Binary Search Balanced Binary Expression Height Balanced Weight Balanced Height Balanced Weight Balanced
- 27. Tuesday, April 8, 2014 27 Trees Data Structures • Tree – Nodes – Each node can have 0 or more children – A node can have at most one parent • Binary tree – Tree with 0–2 children per node Tree Binary Tree
- 28. Tuesday, April 8, 2014 28 Trees • Terminology – Root no parent – Leaf no child – Interior non-leaf – Height distance from root to leaf Root node Leaf nodes Interior nodes Height
- 29. Tuesday, April 8, 2014 29 The degree (shown in green color)of a tree is the maximum degree of node in a tree. Depth (shown in red color) of a tree is the maximum level of any node in a tree. K L E F B G C M H I J D A Level 1 2 3 4 3 2 1 3 2 0 0 1 0 0 0 0 0 1 2 2 2 3 3 3 3 3 3 4 4 4
- 30. Tuesday, April 8, 2014 30 Binary Trees A binary tree is a tree in which no node can have more than two child nodes (left and right child). 2-tree or extended binary tree: Each node has either no children or two children. It is used to represent and compute any algebraic expression using binary operation. e.g. E = (a + b) / ((c - d) * e) / + * a b - e c d Fig. - expression tree
- 31. Tuesday, April 8, 2014 31 Samples of Trees A B A B A B C GE I D H F Complete Binary Tree Skewed Binary Tree E C D 1 2 3 4 5
- 32. Tuesday, April 8, 2014 32 Tree Traversal 1. Preorder Traversal (Root -> Left -> Right) 2. In order Traversal (Left -> Root -> Right) 3. Post order traversal (Left -> Right -> Root) In order traversal – prints the node values in ascending order • Traverse the left sub tree with an in order traversal • Process the value in the node (i.e., print the node value) • Traverse the right sub tree with an in order traversal Preorder traversal •Process the value in the node •Traverse the left sub tree with a preorder traversal •Traverse the right sub tree with a preorder traversal Post order traversal •Traverse the left sub tree with a post order traversal •Traverse the right sub tree with a post order traversal •Process the value in the node
- 33. Tuesday, April 8, 2014 33 preorder: A B C D E F G H I inorder: B C A E D G H F I A B, C D, E, F, G, H, I A D, E, F, G, H, IB C A B C D E F G I H
- 34. Tuesday, April 8, 2014 34 1 /* tree.c 2 Create a binary tree and traverse it 3 preorder, inorder, and postorder */ 4 #include <stdio.h> 5 #include <stdlib.h> 6 #include <time.h> 7 8 struct treeNode { 9 struct treeNode *leftPtr; 10 int data; 11 struct treeNode *rightPtr; 12 }; 13 14 typedef struct treeNode TreeNode; 15 typedef TreeNode *TreeNodePtr; 16 17 void insertNode( TreeNodePtr *, int ); 18 void inOrder( TreeNodePtr ); 19 void preOrder( TreeNodePtr ); 20 void postOrder( TreeNodePtr ); 21 22 int main() 23 { 24 int i, item; 25 TreeNodePtr rootPtr = NULL; 26 27 srand( time( NULL ) ); 28 Outline 1. Define struct 1.1 Function prototypes 1.2 Initialize variables
- 35. Tuesday, April 8, 2014 35 Outline 1.3 Insert random elements 2. Function calls 3. Function definitions 29 /* insert random values between 1 and 15 in the tree */ 30 printf( "The numbers being placed in the tree are:n" ); 31 32 for ( i = 1; i <= 10; i++ ) { 33 item = rand() % 15; 34 printf( "%3d", item ); 35 insertNode( &rootPtr, item ); 36 } 37 38 /* traverse the tree preOrder */ 39 printf( "nnThe preOrder traversal is:n" ); 40 preOrder( rootPtr ); 41 42 /* traverse the tree inOrder */ 43 printf( "nnThe inOrder traversal is:n" ); 44 inOrder( rootPtr ); 45 46 /* traverse the tree postOrder */ 47 printf( "nnThe postOrder traversal is:n" ); 48 postOrder( rootPtr ); 49 50 return 0; 51 } 52 53 void insertNode( TreeNodePtr *treePtr, int value ) 54 { 55 if ( *treePtr == NULL ) { /* *treePtr is NULL */ 56 *treePtr = malloc( sizeof( TreeNode ) ); 57 58 if ( *treePtr != NULL ) { 59 ( *treePtr )->data = value; 60 ( *treePtr )->leftPtr = NULL; 61 ( *treePtr )->rightPtr = NULL; 62 }
- 36. Tuesday, April 8, 2014 36 Outline 3. Function definitions 63 else 64 printf( "%d not inserted. No memory available.n", 65 value ); 66 } 67 else 68 if ( value < ( *treePtr )->data ) 69 insertNode( &( ( *treePtr )->leftPtr ), value ); 70 else if ( value > ( *treePtr )->data ) 71 insertNode( &( ( *treePtr )->rightPtr ), value ); 72 else 73 printf( "dup" ); 74 } 75 76 void inOrder( TreeNodePtr treePtr ) 77 { 78 if ( treePtr != NULL ) { 79 inOrder( treePtr->leftPtr ); 80 printf( "%3d", treePtr->data ); 81 inOrder( treePtr->rightPtr ); 82 } 83 } 84 85 void preOrder( TreeNodePtr treePtr ) 86 { 87 if ( treePtr != NULL ) { 88 printf( "%3d", treePtr->data ); 89 preOrder( treePtr->leftPtr ); 90 preOrder( treePtr->rightPtr ); 91 } 92 }
- 37. Tuesday, April 8, 2014 37 Outline 3. Function definitions Program Output 93 94 void postOrder( TreeNodePtr treePtr ) 95 { 96 if ( treePtr != NULL ) { 97 postOrder( treePtr->leftPtr ); 98 postOrder( treePtr->rightPtr ); 99 printf( "%3d", treePtr->data ); 100 } 101 } The numbers being placed in the tree are: 7 8 0 6 14 1 0dup 13 0dup 7dup The preOrder traversal is: 7 0 6 1 8 14 13 The inOrder traversal is: 0 1 6 7 8 13 14 The postOrder traversal is: 1 6 0 13 14 8 7
- 38. Tuesday, April 8, 2014 38 Binary Search Trees • Key property – Value at node • Smaller values in left subtree • Larger values in right subtree – Example • X > Y • X < Z Y X Z
- 39. Tuesday, April 8, 2014 39 Binary Search Trees Binary search tree • Values in left sub tree less than parent • Values in right sub tree greater than parent • Applications: Facilitates duplicate elimination, generating Huffman codes, expression tree • Fast searches - for a balanced tree, maximum of log n comparisons 47 25 77 11 43 65 93 687 17 31 44
- 40. Tuesday, April 8, 2014 40 Building expression tree: An Application a b c * + Step 1: Push a, b and c sub trees into stack. Step 2: When operator * is encountered, pop top two sub trees from stack and build tree as: * b c Step 3: Push the above sub tree onto stack. Step 4: When operator + is encountered, pop top two sub trees from stack and build the tree as: + a * b c
- 41. Tuesday, April 8, 2014 41 Binary Search Trees • Examples Binary search trees Not a binary search tree 5 10 30 2 25 45 5 10 45 2 25 30 5 10 30 2 25 45
- 42. Tuesday, April 8, 2014 42 Binary Tree Implementation Class Node { int data; // Could be int, a class, etc Node *left, *right; // null if empty void insert ( int data ) { … } void delete ( int data ) { … } Node *find ( int data ) { … } … }
- 43. Tuesday, April 8, 2014 43 Iterative Search of Binary Tree Node *Find( Node *n, int key) { while (n != NULL) { if (n->data == key) // Found it return n; if (n->data > key) // In left subtree n = n->left; else // In right subtree n = n->right; } return null; } Node * n = Find( root, 5);
- 44. Tuesday, April 8, 2014 44 Recursive Search of Binary Tree Node *Find( Node *n, int key) { if (n == NULL) // Not found return( n ); else if (n->data == key) // Found it return( n ); else if (n->data > key) // In left subtree return Find( n->left, key ); else // In right subtree return Find( n->right, key ); } Node * n = Find( root, 5);
- 45. Tuesday, April 8, 2014 45 Example Binary Searches • Find ( root, 2 ) 5 10 30 2 25 45 5 10 30 2 25 45 10 > 2, left 5 > 2, left 2 = 2, found 5 > 2, left 2 = 2, found root
- 46. Tuesday, April 8, 2014 46 Example Binary Searches • Find (root, 25 ) 5 10 30 2 25 45 5 10 30 2 25 45 10 < 25, right 30 > 25, left 25 = 25, found 5 < 25, right 45 > 25, left 30 > 25, left 10 < 25, right 25 = 25, found
- 47. Tuesday, April 8, 2014 47 Binary Search Tree – Insertion • Algorithm 1. Perform search for value X 2. Search will end at node Y (if X not in tree) 3. If X < Y, insert new leaf X as new left subtree for Y 4. If X > Y, insert new leaf X as new right subtree for Y • Observations – O( log(n) ) operation for balanced tree – Insertions may unbalance tree
- 48. Tuesday, April 8, 2014 48 Example Insertion • Insert ( 20 ) 5 10 30 2 25 45 10 < 20, right 30 > 20, left 25 > 20, left Insert 20 on left 20
- 49. Tuesday, April 8, 2014 49 Insert Node in Binary Search Tree 40 30 5 2 30 5 40 2 35 80 Insert 80 Insert 35 30 5 40 2 80
- 50. Tuesday, April 8, 2014 50 Binary Search Tree – Deletion • Algorithm 1. Perform search for value X 2. If X is a leaf, delete X 3. Else // must delete internal node a) Replace with largest value Y on left subtree OR smallest value Z on right subtree b) Delete replacement value (Y or Z) from subtree Observation – O( log(n) ) operation for balanced tree – Deletions may unbalance tree
- 51. Tuesday, April 8, 2014 51 Example Deletion (Leaf) • Delete ( 25 ) 5 10 30 2 25 45 10 < 25, right 30 > 25, left 25 = 25, delete 5 10 30 2 45
- 52. Tuesday, April 8, 2014 52 Example Deletion (Internal Node) • Delete ( 10 ) 5 10 30 2 25 45 5 5 30 2 25 45 2 5 30 2 25 45 Replacing 10 with largest value in left subtree Replacing 5 with largest value in left subtree Deleting leaf
- 53. Tuesday, April 8, 2014 53 Example Deletion (Internal Node) • Delete ( 10 ) 5 10 30 2 25 45 5 25 30 2 25 45 5 25 30 2 45 Replacing 10 with smallest value in right subtree Deleting leaf Resulting tree
- 54. Tuesday, April 8, 2014 54 Deletion for A Binary Search Tree 40 20 60 10 30 50 70 45 55 52 40 20 55 10 30 50 70 45 52 Before deleting 60 After deleting 60 non-leaf node
- 55. Tuesday, April 8, 2014 55 Binary Search Properties • Time of search – Proportional to height of tree – Balanced binary tree • O( log(n) ) time – Degenerate tree • O( n ) time • Like searching linked list / unsorted array
- 56. Tuesday, April 8, 2014 56 AVL tree AVL trees are height-balanced binary search trees Balance factor of a node= height(left sub tree) - height(right sub tree) An AVL tree has balance factor calculated at every node For every node, heights of left and right sub tree can differ by no more than 1 Store current heights in each node AVL property violated here
- 57. Tuesday, April 8, 2014 57 Rotations • When the tree structure changes (e.g., insertion or deletion), we need to transform the tree to restore the AVL tree property. • This is done using single rotations or double rotations. x y A B C y x A B C Before Rotation After Rotation e.g. Single Rotation
- 58. Tuesday, April 8, 2014 58 Rotations • Since an insertion/deletion involves adding/deleting a single node, this can only increase/decrease the height of some subtree by 1 • Thus, if the AVL tree property is violated at a node x, it means that the heights of left(x) and right(x) differ by exactly 2. • Rotations will be applied to x to restore the AVL tree property.
- 59. Tuesday, April 8, 2014 59 Single rotation The new key is inserted in the subtree A. The AVL-property is violated at x height of left(x) is h+2 height of right(x) is h.
- 60. Tuesday, April 8, 2014 60 Single rotation Single rotation takes O(1) time. Insertion takes O(log N) time. The new key is inserted in the subtree C. The AVL-property is violated at x.
- 61. Tuesday, April 8, 2014 61 5 3 1 4 Insert 0.8 AVL Tree 8 0.8 5 3 1 4 8 x y A B C 3 51 0.8 4 8 After rotation
- 62. Tuesday, April 8, 2014 62 Double rotation The new key is inserted in the subtree B1 or B2. The AVL-property is violated at x. x-y-z forms a zig-zag shape also called left-right rotate
- 63. Tuesday, April 8, 2014 63 Double rotation The new key is inserted in the subtree B1 or B2. The AVL-property is violated at x. also called right-left rotate
- 64. Tuesday, April 8, 2014 64 5 3 1 4 Insert 3.5 AVL Tree 8 3.5 5 3 1 4 8 4 5 1 3 3.5 After Rotation x y A z B C 8
- 65. Tuesday, April 8, 2014 65 An Extended Example Insert 3,2,1,4,5,6,7, 16,15,14 3 Fig 1 3 2 Fig 2 3 2 1 Fig 3 2 1 3 Fig 4 2 1 3 4 Fig 5 2 1 3 4 5 Fig 6 Single rotation Single rotation
- 66. Tuesday, April 8, 2014 66 Insert 3,2,1,4,5,6,7, 16,15,14 2 1 4 53 Fig 7 6 2 1 4 53 Fig 8 4 2 5 61 3 Fig 9 4 2 5 61 3 7Fig 10 4 2 6 71 3 5 Fig 11 Single rotation Single rotation
- 67. Tuesday, April 8, 2014 67 4 2 6 71 3 5 16 Fig 12 4 2 6 71 3 5 16 15 Fig 13 4 2 6 151 3 5 16 7Fig 14 Double rotation Insert 3,2,1,4,5,6,7, 16,15,14
- 68. Tuesday, April 8, 2014 68 5 4 2 7 151 3 6 1614 Fig 16 4 2 6 151 3 5 16 7 14 Fig 15 Double rotation Insert 3,2,1,4,5,6,7, 16,15,14
- 69. Tuesday, April 8, 2014 69 j k X Y Z Consider a valid AVL subtree AVL Insertion: Outside Case h h h
- 70. Tuesday, April 8, 2014 70 j k X Y Z Inserting into X destroys the AVL property at node j AVL Insertion: Outside Case h h+1 h
- 71. Tuesday, April 8, 2014 71 j k X Y Z Do a “right rotation” AVL Insertion: Outside Case h h+1 h
- 72. Tuesday, April 8, 2014 72 j k X Y Z Do a “right rotation” Single right rotation h h+1 h
- 73. Tuesday, April 8, 2014 73 j k X Y Z “Right rotation” done! (“Left rotation” is mirror symmetric) Outside Case Completed AVL property has been restored! h h+1 h
- 74. Tuesday, April 8, 2014 74 j k X Y Z AVL Insertion: Inside Case Consider a valid AVL subtree h hh
- 75. Tuesday, April 8, 2014 75 Inserting into Y destroys the AVL property at node j j k X Y Z AVL Insertion: Inside Case Does “right rotation” restore balance? h h+1h
- 76. Tuesday, April 8, 2014 76 j k X Y Z “Right rotation” does not restore balance… now k is out of balance AVL Insertion: Inside Case h h+1 h
- 77. Tuesday, April 8, 2014 77 Consider the structure of subtree Y… j k X Y Z AVL Insertion: Inside Case h h+1h
- 78. Tuesday, April 8, 2014 78 j k X V Z W i Y = node i and subtrees V and W AVL Insertion: Inside Case h h+1h h or h-1
- 79. Tuesday, April 8, 2014 79 j k X V Z W i AVL Insertion: Inside Case We will do a left-right “double rotation” . . .
- 80. Tuesday, April 8, 2014 80 j k X V Z W i Double rotation : first rotation left rotation complete
- 81. Tuesday, April 8, 2014 81 j k X V Z W i Double rotation : second rotation Now do a right rotation
- 82. Tuesday, April 8, 2014 82 jk X V ZW i Double rotation : second rotation right rotation complete Balance has been restored hh h or h-1
- 83. Tuesday, April 8, 2014 83 Arguments for AVL trees: 1. Search is O(log N) since AVL trees are always balanced. 2. Insertion and deletions are also O(logn) 3. The height balancing adds no more than a constant factor to the speed of insertion. Arguments against using AVL trees: 1. Difficult to program & debug; more space for balance factor. 2. Asymptotically faster but rebalancing costs time. 3. Most large searches are done in database systems on disk and use other structures (e.g. B-trees). 4. May be OK to have O(N) for a single operation if total run time for many consecutive operations is fast (e.g. Splay trees). Pros and Cons of AVL Trees
- 84. Tuesday, April 8, 2014 84 All leaves are on the bottom level. All internal nodes (except perhaps the root node) have at least ceil(m / 2) (nonempty) children. The root node can have as few as 2 children if it is an internal node, and can obviously have no children if the root node is a leaf (that is, the whole tree consists only of the root node). Each leaf node (other than the root node if it is a leaf) must contain at least ceil(m / 2) - 1 keys. B-tree is a fairly well-balanced tree. B-Tree
- 85. Tuesday, April 8, 2014 85 Let's work our way through an example similar to that given by Kruse. Insert the following letters into what is originally an empty B-tree of order 5: C N G A H E K Q M F W L T Z D P R X Y S Order 5 means that a node can have a maximum of 5 children and 4 keys. All nodes other than the root must have a minimum of 2 keys. The first 4 letters get inserted into the same node, resulting in this picture:
- 86. Tuesday, April 8, 2014 86 When we try to insert the H, we find no room in this node, so we split it into 2 nodes, moving the median item G up into a new root node. Note that in practice we just leave the A and C in the current node and place the H and N into a new node to the right of the old one. Inserting E, K, and Q proceeds without requiring any splits: H E K Q M F W L T Z D P R X Y S
- 87. Tuesday, April 8, 2014 87 Inserting M requires a split. Note that M happens to be the median key and so is moved up into the parent node. The letters F, W, L, and T are then added without needing any split. M F W L T Z D P R X Y S
- 88. Tuesday, April 8, 2014 88 When Z is added, the rightmost leaf must be split. The median item T is moved up into the parent node. Note that by moving up the median key, the tree is kept fairly balanced, with 2 keys in each of the resulting nodes. The insertion of D causes the leftmost leaf to be split. D happens to be the median key and so is the one moved up into the parent node. The letters P, R, X, and Y are then added without any need of splitting: Z D P R X Y S
- 89. Tuesday, April 8, 2014 89 Finally, when S is added, the node with N, P, Q, and R splits, sending the median Q up to the parent. However, the parent node is full, so it splits, sending the median M up to form a new root node. Note how the 3 pointers from the old parent node stay in the revised node that contains D and G. S
- 90. HEAP A max tree is a tree in which the key value in each node is no smaller than the key values in its children. A max heap is a complete binary tree that is also a max tree. A min tree is a tree in which the key value in each node is no larger than the key values in its children. A min heap is a complete binary tree that is also a min tree. Operations on heaps: - creation of an empty heap - insertion of a new element into the heap; - deletion of the largest element from the heap Tuesday, April 8, 2014 90
- 91. Tuesday, April 8, 2014 91 14 12 7 810 6 9 6 3 5 30 25 [1] [2] [3] [5] [6] [1] [2] [3] [4] [1] [2] 2 7 4 810 6 10 20 83 50 11 21 [1] [2] [3] [5] [6] [1] [2] [3] [4] [1] [2] [4] Max Heap Min Heap
- 92. Tuesday, April 8, 2014 92 Example of Insertion to Max Heap 20 15 2 14 10 initial location of new node 21 15 20 14 10 2 insert 21 into heap 20 15 5 14 10 2 insert 5 into heap
- 93. Tuesday, April 8, 2014 93 Insertion into a Max Heap void insert_max_heap(element item, int *n) { int i; if (HEAP_FULL(*n)) { fprintf(stderr, “the heap is full.n”); exit(1); } i = ++(*n); while ((i!=1)&&(item.key>heap[i/2].key)) { heap[i] = heap[i/2]; i /= 2; } heap[i]= item; } 2k-1=n ==> k=log2(n+1) O(log2n)
- 94. Tuesday, April 8, 2014 94 Example of Deletion from Max Heap 20 remove 15 2 14 10 10 15 2 14 15 14 2 10
- 95. Tuesday, April 8, 2014 95 Deletion from a Max Heap element delete_max_heap(int *n) { int parent, child; element item, temp; if (HEAP_EMPTY(*n)) { fprintf(stderr, “The heap is emptyn”); exit(1); } /* save value of the element with the highest key */ item = heap[1]; /* use last element in heap to adjust heap */ temp = heap[(*n)--]; parent = 1; child = 2;
- 96. Tuesday, April 8, 2014 96 while (child <= *n) { /* find the larger child of the current parent */ if ((child < *n)&& (heap[child].key<heap[child+1].key)) child++; if (temp.key >= heap[child].key) break; /* move to the next lower level */ heap[parent] = heap[child]; child *= 2; } heap[parent] = temp; return item; }
- 97. Tuesday, April 8, 2014 97 Thank You Contact @ Email: kuberchandra@yahoo.com

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