## 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 = 2Output:[2,1,3]

**Example 2:**

Input:root = [4,2,7,1,3], val = 5Output:[]

**Constraints:**

- The number of nodes in the tree is in the range
`[1, 5000]`

. `1 <= Node.val <= 10`

^{7}`root`

is a binary search tree.`1 <= val <= 10`

^{7}

## 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