700-search-in-a-binary-search-tree

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: []

My solution：

# 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(object):
    def searchBST(self, root, val):
        """
        :type root: Optional[TreeNode]
        :type val: int
        :rtype: Optional[TreeNode]
        """
        if not root:
            return None
        if root.val == val:
            return root
        else:
            if root.val > val:
                return self.searchBST(root.left, val)
            else:
                return self.searchBST(root.right, val)