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.

　　思路：中序遍历的结果正好是二分查找树对应的数据的顺序，所以先中序遍历数据，然后检查遍历的结果是不是递增的

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool check(vector<int> &a)
    {
        if(a.size() == 1) return true;
        for(int i = 1; i< a.size(); i++)
          if(a[i] <= a[i-1])
                return false;
        return true;
   }
    bool isValidBST(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(root == NULL ) return true ;
        vector<int> inorder ;
        stack<TreeNode *> myS;
       // myS.push(root)；
        TreeNode *p = root;
        while(!myS.empty()||p){
           while(p){
             myS.push(p);
             p = p->left;
           }
           p = myS.top();myS.pop();
           inorder.push_back(p->val) ;
           p = p->right ;
        }
        
        return check(inorder) ;
        
    }
};

 改进版本：访问每个数据的时候就进行检查，奇怪的是小数据时时间减小一半，大数据竟然比上面的用时还要多

 bool isValidBST(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(root == NULL ) return true ;

        stack<TreeNode *> myS;
       // myS.push(root)；
        TreeNode *p = root;
        int flag = 0;
        int value ;
        while(!myS.empty()||p){
           while(p){
             myS.push(p);
             p = p->left;
           }
           p = myS.top();
           myS.pop();
          
           if(flag){
               
              if( p->val <= value)
                  return false;
               
           }else{
               
               flag = 1;
             
           }
           value  = p->val;
           p = p->right ;
        }
        return true ;
        
    }