16. Binary Search Tree - BinarySearchTree 1...

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Binary Search Tree 1
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2 Binary Search Tree (BST)  - It is a hierarchical, nonlinear data structure.  - Its properties are given bellow:     - Each node has a value.     - A total order is defined on these node values.     - Left subtree of a node contains the values less than the node’s value.     - Right subtree of a node contains the values greater than or equal to        the node’s value. Example: Root R 8 3 12 1 6 9 14 2 10 Left Subtree Right Subtree
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3 Searching a node with a specific value from BST Rules Examine the root R. If the searching value equals the root, then the value exists in the tree. If it is less than the root, then it must be in the left subtree. If it is greater than the root, then it must be in the right subtree. This process continues until the searching value is found or a leaf node is reached without having that value where it would be. Algorithm Search_BST(Root, Left, Right, Key ) 1. Loc := Search(Root, Key, NULL) 2. If Loc = NULL then Print: “Item not in BST”. 3. Else Print: “Item is in BST”. 4. End. Here, Root = Root of BST Left = Left Subtree Right = Right Subtree Key = Searching Item
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4 Search(Node, Key, Par) 1. If Node = NULL then return NULL. 2. If Key < Node.Key then Search(Node.Left, Key, Node). 3. Else if Key > Node.Key then Search(Node.Right, Key, Node). 4. Else return Node Example: Searching Item:  12 Steps: 1. Search(Root, 12)                Search(AA,12) 2. Search(AA->Left,12)          Search(BB, 12) 3. Search(BB->Left, 12)         Search(DD, 12) 4. Return DD and Print: Item is in BST. 15 13 100 12 90 AA BB CC DD EE
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5 Complexity of Searching a Node from BST  In Binary Search Tree, Depth = log 2 N  For balanced BST, Complexity of Searching=O(log 2 N)  [Average Case]   For unbalanced BST, Complexity of Searching=O(N)     [Worst Case]
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