LeetCode #98: Validate Binary Search Tree

Description:

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:

    2
   / \
  1   3

Input: [2,1,3]
Output: true

Example 2:

    5
   / \
  1   4
     / \
    3   6

Input: [5,1,4,null,null,3,6]
Output: false
Explanation: The root node's value is 5 but its right child's value is 4.

Solution:

• Traverse the tree recursively, if both left and right child trees are BST and the root value in in between the maximum value of left child tree and the minimum value of right child tree, it is a valid BST.
• Time complexity: O(N)
• Space complexity: O(1)

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isValidBST(self, root: TreeNode) -> bool:
        xmax = float('-inf')
        flag, xmax = self.inorder(root, xmax)
        return flag


    def inorder(self, node, xmax):
        if node is not None:
            #print(node.val, xmax)
            left, xmax = self.inorder(node.left, xmax)
            if left and (node.val > xmax):
                xmax = node.val
            else:
                return False, xmax
            return self.inorder(node.right, xmax)
        else:
            return True, xmax