Search in a Binary Search Tree - Tree - Easy - LeetCode
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Search in a Binary Search Tree - Tree - Easy - LeetCode

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  • 1To find a node in a binary search tree (BST), compare the node's value with the target value recursively.
  • 2If the target value matches a node's value, return the subtree rooted at that node; otherwise, return NULL.
  • 3The search operation has a time complexity of O(log n) and a space complexity of O(1) in a balanced BST.

AI-generated summary · May not capture all nuances

Key Insight
AskGif

"To find a node in a binary search tree (BST), compare the node's value with the target value recursively."

Search in a Binary Search Tree - Tree - Easy - LeetCode

Given the root node of a binary search tree (BST) and a value. You need to find the node in the BST that the node's value equals the given value. Return the subtree rooted with that node. If such node doesn't exist, you should return NULL.

For example,

Given the tree: 4 / \ 2 7 / \ 1 3

And the value to search: 2 You should return this subtree:

2 / \ 1 3 In the example above, if we want to search the value 5, since there is no node with value 5, we should return NULL.

Note that an empty tree is represented by NULL, therefore you would see the expected output (serialized tree format) as [], not null.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 * public int val;
 * public TreeNode left;
 * public TreeNode right;
 * public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 * this.val = val;
 * this.left = left;
 * this.right = right;
 * }
 * }
 */
public class Solution {
 public TreeNode SearchBST(TreeNode root, int val) {
 if(root == null){
 return null;
 }
 
 if(root.val>val){
 return SearchBST(root.left,val);
 }
 else if(root.val<val){
 return SearchBST(root.right,val);
 }
 else if(root.val==val){
 return root;
 }
 
 return null;
 }
}

Time Complexity: O(logn)

Space Complexity: O(1)

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sumitc91

Published on 9 October 2020 · 1 min read · 215 words

Part of AskGif Blog · coding

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