LeetCode98：Validate Binary Search Tree

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.

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LeetCode：链接

题目要求去validate一个tree是不是Binary Search Tree，也就是所有left subtree的值都比现在node的值小，所有right subtree的值都比现在node的值大。

要注意的坑就是有可能是形如：

这就不是有效的binary search tree了，因为3虽然比8小，但是必须要比5大，所以在写代码的时候不能只比较当前父节点的大小，还要考虑之前所有的父节点情况。所以我们必须要有两个变量，lowerbound和upperbound。每次递归的时候更新这两个变量。

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

class Solution(object):
    def isValidBST(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        lowerbound = float('-inf')
        upperbound = float('inf')
        return self.ValidBST(root, lowerbound, upperbound)

    def ValidBST(self, root, lowerbound, upperbound):
        if not root:
            return True
        if root.val <= lowerbound or root.val >= upperbound:
            return False
        return self.ValidBST(root.left, lowerbound, root.val) and self.ValidBST(root.right, root.val, upperbound)