[LeetCode] Validate Binary Search Tree

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.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

递归判断，递归时传入两个参数，一个是左界，一个是右界，节点的值必须在两个界的中间，同时在判断做子树和右子树时更新左右界。

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     bool check(TreeNode *node, int leftVal, int rightVal)
13     {
14         if (node == NULL)
15             return true;
16             
17         return leftVal < node->val && node->val < rightVal && check(node->left, leftVal, node->val) &&
18             check(node->right, node->val, rightVal);
19     }
20     
21     bool isValidBST(TreeNode *root) {
22         // Start typing your C/C++ solution below
23         // DO NOT write int main() function
24         return check(root, INT_MIN, INT_MAX);        
25     }
26 };