Leetcode 98. Validate Binary Search Tree

Description:Given the root of a binary tree, determine if it is a valid binary search tree (BST).

A valid 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.

Link: 98. Validate Binary Search Tree

Examples:

Example 1:

Input: root = [2,1,3]
Output: true

Example 2:

Input: root = [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.

思路: 做过了上面的那道为二叉搜索树换位置，觉得这个好简单，却是机关重重。首先我们会想，记录前一个节点pre，然后值大于当前就返回False,然后会发现错了，为什么？因为这是在递归里返回False,而没有彻底从函数栈中出来，所以还是要记录遍历的过程有没有过False的情况，如果有，在总函数中返回False. 但是[1,1]这个case没过，所以条件不是大于，是大于等于，然后在过了。

class Solution(object):
    def isValidBST(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        self.pre = None
        self.a = []
        self.Inorder(root)
        if self.a:
            return False
        return True
        
    def Inorder(self, root):
        if not root: return
        self.Inorder(root.left)
        if self.pre and self.pre.val >= root.val:
            self.a.append(False)
        self.pre = root
        self.Inorder(root.right)

日期: 2021-03-16