Tags: "leetcode", "bfs", "bst", access_time 2-min read

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Validate Binary Search Tree

Created: November 10, 2018 by [lek-tin]

Last updated: November 10, 2018

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

  • The left subtree of a node contains only nodes with keys less than the node’s key.
  • The right subtree of a node contains only nodes with keys greater than the node’s key.
  • Both the left and right subtrees must also be binary search trees.

Example 1

Input:
    2
   / \
  1   3
Output: true

Example 2

    5
   / \
  1   4
     / \
    3   6
Output: false

Explanation The input is: [5,1,4,null,null,3,6]. The root node’s value is 5 but its right child’s value is 4.

Solution

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
# Time - O(n)
# Space - O(n)
class Solution:
    def isValidBST(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        # root is None;
        if not root:
            return True
        return self.validate(root, None, None)

    def validate(self, node, minVal, maxVal):
        # Reaches leftmost/rightmost node
        if not node:
            return True
        # validate left node
        if maxVal != None and node.val >= maxVal:
            return False
        # validate right node
        if minVal != None and node.val <= minVal:
            return False

        return self.validate(node.left, minVal, node.val) and self.validate(node.right, node.val, maxVal)