# [Solved] 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.

Table of Contents

## Question

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`

## Python Solution

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