[leetcode] 98. Validate Binary Search Tree

原创

是念 2022-08-12 08:26:43 博主文章分类：C++ ©著作权

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Description

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

Input:

[2,1,3]

Output:

true

Example 2:

Input:

[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.

分析

题目的意思是：判断一颗二叉树是否是合法的二叉搜索树。

对于一棵二叉搜索树，根结点必须比所有的左子树的值大，然后比右子树的所有的值小。这样就很好理解为什么要用min,max指针了。
注意数的溢出问题。

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) {
        return judge(root,LONG_MIN,LONG_MAX);
    }
    
    bool judge(TreeNode *root, long min_val,long max_val){
        if(root==NULL){
            return true;
        }
        if(root->val<=min_val||root->val>=max_val){
            return false;
        }
        return judge(root->left,min_val,root->val)&&judge(root->right,root->val,max_val);
    }
};