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Basic introduction on how to use Binary Search Tree and AVL.

Basic introduction on how to use Binary Search Tree and AVL.

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- 1. Binary Search Tree
- 2. PreliminariesSearch Array List Stack? Queue? For large amounts of input, the linear access time of lists is prohibitive
- 3. PreliminariesTree A collection of nodes The collection can be empty Otherwise, the tree consists of a distinguished node r, called the root, and zero or more (sub)trees T1, T2, T3, …, Tk, each of whose roots are connected by a directed edge to r. The root of each subtree is said to be a child of r, and r is the parent of each subtree root
- 4. Preliminaries rootT1 T2 T3 T4 Tk …
- 5. PreliminariesTree A tree is a collection of n nodes, one of which is the root, and n-1 edges. Each edge connects some node to its parent and every node except the root has one parent
- 6. Preliminaries ROOT A INTERNAL NODES INTERNAL NODES B C D E F GLEAF LEAF H I J L M N O LEAF LEAF LEAVES P Q LEAVES
- 7. Binary TreeA binary tree is a tree in which no node can have more than two children root TL TR
- 8. Binary Tree A D EH I J L P Q
- 9. Implementation Nodeclass node{public: int item; node *left; node *right; node(int x) { item = x; left = right = NULL; } node( ) { item = 0; left = right = NULL; }};
- 10. Expression Trees(a+b*c)+((d*e+f)*g) + + *a * + g b c * f d e
- 11. Binary Search Tree An application of binary trees is their use in searching Let us assume that each node in the tree is assigned a key value, and assume that this is an integer The property that makes a binary tree into a binary search tree is that for every node, X, in the tree, the values of all keys in the left subtree are smaller than the key value in X, and the values of all keys in the right subtree are larger than the key value in X.
- 12. Binary Search Tree 6 6 2 8 2 81 4 1 4 3 3 7
- 13. Binary Search Treeclass BST{private: int size; node *root;public: BST() {size = 0; root = NULL;} void insert(int); bool delete(int); bool search(int); int minimum(); int maximum();};
- 14. Binary Search Tree (Search) Search for 10 6 2 8 1 4 20 3 10 50
- 15. Binary Search Tree (Search)bool BST::search(int x){ node *tmp = root; while(tmp!=NULL){ if(x == tmp->item) return true; if(x < tmp->item) tmp = tmp->left; else tmp = tmp->right; } return false;}
- 16. Binary Search Tree (Minimum) 6 2 8 1 4 20 -5 3 10 50 -1
- 17. Binary Search Tree (Minimum)int BST::minimum(){ node *tmp = root; while(tmp->left != NULL) tmp = tmp -> left; return temp->item;}
- 18. Binary Search Tree (Insert) 6Insert 20, 10, 50 2 8 1 4 20 3 10 50
- 19. Binary Search Tree (Insert)Let’s insert 6, 2, 4, 3, 1, 8 and 11 in an empty BST 6 2 8 1 4 11 3
- 20. Binary Search Tree (Insert)Try inserting 1, 2, 3, and 4 in an empty BST. 1 2 3 4
- 21. Binary Search Tree (Insert)void BST::insert(int x){ node *n = new node(x); node *tmp = root; if(tmp = NULL) root = n; else{ node *tmp2; while(tmp!=NULL){ tmp2 = tmp; if(x < tmp->item) tmp = tmp->left; else tmp = tmp->right; } if(x < tmp2->item) tmp2->left = n; else tmp2->right = n; }}
- 22. BST (Delete)In deletion, we don’t ask for a position. We ask for the actual item that has to be deleted. Deleting a leaf 6 2 8 4 11 1 Deleting a node with one child 3 Deleting a node with two children
- 23. Deleting a Leaf (-1) 6 2 8 1 4 20-5 3 10 50 -1
- 24. Deleting a Node with a Child(-5) 6 2 8 1 4 20 -5 3 10 50 -1
- 25. Deleting a node with two children (2) 6 3 2 8 1 4 20-5 3 10 50 -1
- 26. DeleteNode Codebool BST::deleteNode(int x){ node *del = searchNode(x); if(del->left == NULL && del->right==NULL) delete del; //leaf else{ }}
- 27. One Childif(del->left==NULL) del = del->right;elseif(del->right==NULL) del = del->left;
- 28. Two Childrenelse{ node *ptr = minimum(del->right); int x = ptr->item; deleteNode(x); del->item = x;}
- 29. Running of OperationsLinear 1 2 3 4
- 30. Discussion We did not achieve our goal of log n. Can we improve? Always keep the tree balanced A D E H I J L P Q
- 31. Adelson-Velski Landis (AVL) TreeAn AVL tree is a binary search tree where every node of the tree satisfies the following property: The height of the left and right subtrees of a node differs by at most one.
- 32. Adelson-Velski Landis (AVL) Tree
- 33. Adelson-Velski Landis (AVL) Tree In order to properly implement the AVL, the node has to be redefined.class node{public: int item; node *left; node *right; int height; node(int x) { item = x; left = right = NULL; } node( ) { item = 0; left = right = NULL; }};
- 34. Adelson-Velski Landis (AVL) Tree What kinds of violations may occur in a regular BST that will result in height imbalance?1 3 3 2 2 1 3 1 2
- 35. Right Rotate1 2 2 1 3 3
- 36. Left-Rotate 2 1 3
- 37. Left-Right Rotate 3 21 1 3 2
- 38. Right-Left Rotate3 3 2 2 2 1 31 1
- 39. ChallengeInsert the following items in an AVL 10, 20, 30, 40, 50, 60, 70, 80, 71, 61, 51, 41, 31, 21, 11
- 40. Right Rotatevoid BST::rightRotate(node *r){ node *p = r->left; r->left = p->right; p->right = r; //fill in the missing code}
- 41. Left Rotatevoid BST::leftRotate(node *r){ node *p = r->right; r->right= p->left; p->left= r; //fill in the missing code}
- 42. Other RotationsI leave this as an exercise
- 43. Insertvoid BST::insert(int x){ //do insert as in BST current = x; set current to balanced do{ previous = x; update height of current lh = height of the left subtree of current rh = height of the left subtree of current set current as left leaning, right leaning, or balanced if(violation occurs) perform corresponding rotation }while(??);}

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