[LeetCode] 98. Validate Binary Search Tree

最新推荐文章于 2025-06-18 20:53:42 发布

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本文介绍了如何通过中序遍历判断一棵二叉树是否为有效的二叉搜索树，并给出了具体的Python实现代码。

题：https://leetcode.com/problems/validate-binary-search-tree/description/](https://leetc
ode.com/problems/validate-binary-search-tree/description/)

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.
Example 1:

Input:
2
/ \
1 3
Output: true
Example 2:

/ \
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.

思路

这里用到了一个性质，对于 Binary Search Tree ，它的中序遍历是严格单调递增。

另外，可能使用用动态规划的想法，只要左树的max < 当前节点 < 右树的min，具体的实现还要进一步思考。

code

# 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):
        """
        :type root: TreeNode
        :rtype: bool
        """
        tstack = []
        node = root
        tmp = None
        while node != None or len(tstack) != 0:
            while node != None:
                tstack.append(node)
                node = node.left
            node = tstack.pop(-1)
            print(node.val)
            if tmp != None:
                if not tmp<node.val:
                    return False
            tmp = node.val
            node = node.right
        return True