98.Validate Binary Search Tree（二查搜索树）

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

思想其实很简单，但是我为啥就是想不到呢？？？？？！！！！！

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。

需要考虑结点取INT_MAX 或者INT_MIN的情况，
相应的改成long long 以及 LONG_LONG_MAX 和LONG_LONG_MIN后提交通过。

/**
 * 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 check(TreeNode *node, long long leftVal, long long  rightVal)
     {
         if (node == NULL)
             return true;
             
         return leftVal < node->val && node->val < rightVal && check(node->left, leftVal, node->val) &&
             check(node->right, node->val, rightVal);
     }
     
     bool isValidBST(TreeNode *root) {
         // Start typing your C/C++ solution below
         // DO NOT write int main() function
         return check(root,   LONG_LONG_MIN ,   LONG_LONG_MAX);        
     }
};

　　

2）更好的方法，采用搜索二叉树常用的中序遍历是有序的这个性质，注意代码实现栈的使用，如何防止反复遍历（代码可以套模式的）

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isValidBST(TreeNode* root) {
        if(root==nullptr) {return true;}
        stack<TreeNode*> stk;
        long long inorder = (long long)INT_MIN - 1;

        while(!stk.empty() || root!=nullptr)
        {
            while(root!=nullptr)
            {
                stk.push(root);
                root = root->left;
            }
            root = stk.top();
            stk.pop();
            if(root->val <= inorder) {return false;}
            inorder = root->val; 
            root = root->right;
        }
    return true;
    }
};