Lecture 7-BinarySearchTrees.ppt

CSC803 Data Structures & Algorithm Analysis Binary Search Trees

A Taxonomy of Trees General Trees – any number of children / node Binary Trees – max 2 children / node Heaps – parent < (>) children Binary Search Trees

Binary Trees Binary search tree Every element has a unique key. The keys in a nonempty left subtree (right subtree) are smaller (larger) than the key in the root of subtree. The left and right subtrees are also binary search trees.

Binary Search Trees (BST) are a type of Binary Trees with a special organization of data. This data organization leads to O(log n) complexity for searches, insertions and deletions in certain types of the BST (balanced trees). O(h) in general Binary Search Trees

34 41 56 63 72 89 95 0 1 2 3 4 5 6 34 41 56 0 1 2 72 89 95 4 5 6 34 56 0 2 72 95 4 6 Binary Search algorithm of an array of sorted items reduces the search space by one half after each comparison Binary Search Algorithm

63 41 89 34 56 72 95 • the values in all nodes in the left subtree of a node are less than the node value • the values in all nodes in the right subtree of a node are greater than the node values Organization Rule for BST

Binary Tree typedef struct tnode *ptnode; typedef struct node { short int key; ptnode right, left; } ; sample binary search tree code

Searching in the BST method search(key) • implements the binary search based on comparison of the items in the tree • the items in the BST must be comparable (e.g integers, string, etc.) The search starts at the root. It probes down, comparing the values in each node with the target, till it finds the first item equal to the target. Returns this item or null if there is none. BST Operations: Search

if the tree is empty return NULL else if the item in the node equals the target return the node value else if the item in the node is greater than the target return the result of searching the left subtree else if the item in the node is smaller than the target return the result of searching the right subtree Search in BST - Pseudocode

Search in a BST: C code Ptnode search(ptnode root, int key) { /* return a pointer to the node that contains key. If there is no such node, return NULL */ if (!root) return NULL; if (key == root->key) return root; if (key < root->key) return search(root->left,key); return search(root->right,key); }

method insert(key)  places a new item near the frontier of the BST while retaining its organization of data: starting at the root it probes down the tree till it finds a node whose left or right pointer is empty and is a logical place for the new value uses a binary search to locate the insertion point is based on comparisons of the new item and values of nodes in the BST Elements in nodes must be comparable! BST Operations: Insertion

9 7 5 4 6 8 Case 1: The Tree is Empty Set the root to a new node containing the item Case 2: The Tree is Not Empty Call a recursive helper method to insert the item 10 10 > 7 10 > 9 10

if tree is empty create a root node with the new key else compare key with the top node if key = node key replace the node with the new value else if key > node key compare key with the right subtree: if subtree is empty create a leaf node else add key in right subtree else key < node key compare key with the left subtree: if the subtree is empty create a leaf node else add key to the left subtree Insertion in BST - Pseudocode

Insertion into a BST: C code void insert (ptnode *node, int key) { ptnode ptr, temp = search(*node, key); if (temp || !(*node)) { ptr = (ptnode) malloc(sizeof(tnode)); if (IS_FULL(ptr)) { fprintf(stderr, “The memory is fulln”); exit(1); } ptr->key = key; ptr->left = ptr->right = NULL; if (*node) if (key<temp->key) temp->left=ptr; else temp->right = ptr; else *node = ptr; } }

 The order of supplying the data determines where it is placed in the BST , which determines the shape of the BST  Create BSTs from the same set of data presented each time in a different order: a) 17 4 14 19 15 7 9 3 16 10 b) 9 10 17 4 3 7 14 16 15 19 c) 19 17 16 15 14 10 9 7 4 3 can you guess this shape? BST Shapes

 removes a specified item from the BST and adjusts the tree  uses a binary search to locate the target item:  starting at the root it probes down the tree till it finds the target or reaches a leaf node (target not in the tree) removal of a node must not leave a ‘gap’ in the tree, BST Operations: Removal

method remove (key) I if the tree is empty return false II Attempt to locate the node containing the target using the binary search algorithm if the target is not found return false else the target is found, so remove its node: Case 1: if the node has 2 empty subtrees replace the link in the parent with null Case 2: if the node has a left and a right subtree - replace the node's value with the max value in the left subtree - delete the max node in the left subtree Removal in BST - Pseudocode

Case 3: if the node has no left child - link the parent of the node - to the right (non-empty) subtree Case 4: if the node has no right child - link the parent of the target - to the left (non-empty) subtree Removal in BST - Pseudocode

9 7 5 6 4 8 10 9 7 5 6 8 10 Case 1: removing a node with 2 EMPTY SUBTREES parent cursor Removal in BST: Example Removing 4 replace the link in the parent with null

Case 2: removing a node with 2 SUBTREES 9 7 5 6 8 10 9 6 5 8 10 cursor cursor - replace the node's value with the max value in the left subtree - delete the max node in the left subtree 4 4 Removing 7 Removal in BST: Example What other element can be used as replacement?

9 7 5 6 8 10 9 7 5 6 8 10 cursor cursor parent parent the node has no left child: link the parent of the node to the right (non-empty) subtree Case 3: removing a node with 1 EMPTY SUBTREE Removal in BST: Example

9 7 5 8 10 9 7 5 8 10 cursor cursor parent parent the node has no right child: link the parent of the node to the left (non-empty) subtree Case 4: removing a node with 1 EMPTY SUBTREE Removing 5 4 4 Removal in BST: Example

The complexity of operations get, insert and remove in BST is O(h) , where h is the height. O(log n) when the tree is balanced. The updating operations cause the tree to become unbalanced. The tree can degenerate to a linear shape and the operations will become O (n) Analysis of BST Operations

BST tree = new BST(); tree.insert ("E"); tree.insert ("C"); tree.insert ("D"); tree.insert ("A"); tree.insert ("H"); tree.insert ("F"); tree.insert ("K"); >>>> Items in advantageous order: K H F E D C A Output: Best Case

BST tree = new BST(); for (int i = 1; i <= 8; i++) tree.insert (i); >>>> Items in worst order: 8 7 6 5 4 3 2 1 Output: Worst Case

tree = new BST (); for (int i = 1; i <= 8; i++) tree.insert(random()); >>>> Items in random order: X U P O H F B Output: Random Case

Applications for BST Sorting with binary search trees Input: unsorted array Output: sorted array Algorithm ? Running time ?

Prevent the degeneration of the BST : A BST can be set up to maintain balance during updating operations (insertions and removals) Types of ST which maintain the optimal performance: splay trees AVL trees 2-4 Trees Red-Black trees B-trees Better Search Trees

Lecture 7-BinarySearchTrees.ppt

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Lecture 7-BinarySearchTrees.ppt