【leetcode】98. Validate Binary Search Tree 二叉树是否是中序有序的

1. 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:

2

/ \
1 3
Binary tree [2,1,3], return true.
Example 2:

1

/ \
2 3
Binary tree [1,2,3], return false.

2. 思路

递归测试,满足的条件是左、右子树各自满足,且跟大于左子树max,跟小于又子树min。

3. 代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isValidBST(TreeNode* root) {
        int min, max;
        return isValidBST(root, &min, &max);
    }
    bool isValidBST(TreeNode* root, int* min, int* max) {
        if (root == NULL) { return true; }
        *min = *max = root->val;
        int tmp;
        if (root->left != NULL) {
            if (!isValidBST(root->left, min, &tmp) || tmp >= root->val) { return false; }
        }
        if (root->right != NULL) {
            if (!isValidBST(root->right, &tmp, max) || tmp <= root->val) { return false;}
        }
        return true;
    }
};
    原文作者:knzeus
    原文地址: https://segmentfault.com/a/1190000007445517
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