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  • 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

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  • 原文地址:https://www.cnblogs.com/wangyuxia/p/14544568.html
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