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:

      4
     / \
   (2)  7
  /  \
(1)  (3)

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

Example 2:

     4
    / \
   2   7
  / \
 1   3

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

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# 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 is None or root.val == val:
            return root
        if root.val > val:
            return self.searchBST(root.left, val)
        return self.searchBST(root.right, val)