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.


There are only two possibilities when we searching a node:

  1. node is equal to the value
  2. node is not equal to the value


When the node is equal to the value, we just need to return it. Otherwise, we just need to visit its left or right child recursively.


visit root node.

If root is null or root’s value is equal to target value, return the root.

Otherwises, if root’s value is bigger than target value.

visit root’s left child.


visit root’s right child.


# 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 == None or root.val == val):
            return root
        elif root.val < val:
            return self.searchBST(root.right, val)
            return self.searchBST(root.left, val)

Complexity Analysis

  • Time Complexity: $O(\log{n})$
  • Space Complexity: $O(1)$