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: []

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

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: Optional[TreeNode], val: int) -> Optional[TreeNode]:
        if root:
            if root.val == val:
                return root
            elif val > root.val:
                return self.searchBST(root.right, val)
            else:
                return self.searchBST(root.left, val)
        else:
            return None