In this post, we are going to solve the 700. Search in a Binary Search Tree problem of Leetcode. This problem 700. Search in a Binary Search Tree is a Leetcode easy level problem. Let’s see the code, 700. Search in a Binary Search Tree – Leetcode Solution.

Problem

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.

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

Now, let’s see the code of 700. Search in a Binary Search Tree – Leetcode Solution.

Search in a Binary Search Tree – Leetcode Solution

700. Search in a Binary Search Tree – Solution in Java

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

700. Search in a Binary Search Tree – Solution in C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* searchBST(TreeNode* root, int val) {
        if(!root) return root;

        TreeNode *node=new TreeNode(); 

        if(val<root->val){
            node=searchBST(root->left,val);
        } else if(val>root->val){
            node=searchBST(root->right,val);
        } else {
            node=root;
        }
        return node;
    }
};

700. Search in a Binary Search Tree – Solution in Python

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def searchBST(self, root, val):
        if root and val < root.val: return self.searchBST(root.left, val)
        elif root and val > root.val: return self.searchBST(root.right, val)
        return root
        

Note: This problem 700. Search in a Binary Search Tree is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.