LeetCode 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: []


  • 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


Search recursively base on binary search tree characteristics.

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:
        return self.helper(root, val)
    def helper(self, root, val):
        if not root:
            return None
        if root.val == val:
            return root
        if val < root.val:
            return self.helper(root.left, val)
            return self.helper(root.right, val)
        return None
  • Time Complexity: O(H). H is the tree height.
  • Space Complexity: O(H).

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