[算法分析与设计] leetcode 每周一题: Validate Binary Search Tree

题目链接：https://leetcode.com/problems/validate-binary-search-tree/discuss/

题目：

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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.

Example 1:

    2   / \  1   3

Binary tree [2,1,3], return true.

Example 2:

    1   / \  2   3

Binary tree [1,2,3], return false.

思路：

在二叉搜索树中每一个节点都可能存在约束，都有可能存在一个比该节点大的节点以及一个比该节点小的点，因此用递归，每次传入一个当前结点，极大点，极小点，比较即可。

代码：

class Solution {public:    bool judgeValid(TreeNode * curNode, TreeNode* minNode, TreeNode* maxNode) {       if(!curNode) return true;       if(minNode && minNode->val >= curNode->val || maxNode && maxNode->val <= curNode->val){           return false;       }       return judgeValid(curNode->left, minNode, curNode) && judgeValid(curNode->right, curNode, maxNode);    }    bool isValidBST(TreeNode* root) {        return judgeValid(root, nullptr, nullptr);                   }};