# Search in a Binary Search Tree – Leetcode Solution

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.

Contents

## 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.