98. Validate Binary Search Tree

最新推荐文章于 2020-10-18 10:11:10 发布

原创最新推荐文章于 2020-10-18 10:11:10 发布 · 368 阅读

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#need another try #dfs

本文介绍了一种检查给定二叉树是否为有效二叉搜索树的方法。通过递归方式验证每个节点的左右子树节点值是否符合二叉搜索树的定义，并确保整棵树的结构满足条件。

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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.

/**
 * 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 isValid(TreeNode *root, int maxVal, int minVal, bool checkMax, bool checkMin) {
        if(NULL == root) return true;
        if(checkMax && root->val >= maxVal) return false;
        if(checkMin && root->val <= minVal) return false;
        
return isValid(root->left, root->val, minVal, true, checkMin) && isValid(root->right, maxVal, root->val, checkMax, true);
    
    }
    
    bool isValidBST(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        return isValid(root,0,0,false,false);
    }
};