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 ;

}