[Leetcode] 98. Validate Binary Search Tree @python

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本文介绍了一种判断二叉树是否为有效二叉搜索树的方法。通过递归检查每个节点的值是否在其合法范围内，确保左子树的所有节点值小于根节点值，而右子树的所有节点值大于根节点值。

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题目

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.
confused what “{1,#,2,3}” means? > read more on how binary tree is serialized on OJ.

题目要求

判断一个binary-tree是否是有效的。

解题思路

一个binary-tree有效，要求root节点的左子树的节点的值都小于root节点的值，root节点的右子树的节点的值都大于root节点的值。所以遍历时只要判断节点的值是否处于该子树的允许范围即可。

代码

# 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
        """
        def isValid(root,min_val,max_val):
            if min_val != None and root.val <= min_val:
                return False
            if max_val != None and root.val >= max_val:
                return False
            if root.left:
                if not isValid(root.left,min_val,root.val):
                    return False
            if root.right:
                if not isValid(root.right,root.val,max_val):
                    return False
            return True

        if root == None:
            return True

        return isValid(root,None,None)