LeetCode Link: LC 700 Search in a Binary Search Tree {using Recursion}
Language: C#
Note: This is the same problem from here, but implemented using Recursion.
Problem Statement
You are given the root of a binary search tree (BST) and an integer val.
Find the node in the BST that the node’s value equals val and return the subtree rooted with that node. If such a node does not exist, return null.
Examples
Example 1:
Input: root = [4,2,7,1,3], val = 2
Output: [2,1,3]
Example 2:
Input: root = [4,2,7,1,3], val = 5
Output: []
Constraints
- The number of nodes in the tree is in the range [1, 5000].
- 1 <= Node.val <= 107
- root is a binary search tree.
- 1 <= val <= 107
Solution
/**
* 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 is null || root.val == val)
{
return root;
}
if (val < root.val)
{
return SearchBST(root.left, val);
}
else
{
return SearchBST(root.right, val);
}
}
}
Complexity
Time Complexity: O(N)
Space Complexity: O(1)