[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.
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

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

/**
* 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 isValidBST(TreeNode* root) {
stack<TreeNode*> sta;
TreeNode* T=root;
double left_ele = INT_MIN-1.0;
while(!sta.empty()||T){
while(T){
sta.push(T);
T=T->left;
}
T = sta.top();
sta.pop();
if(T->val>left_ele)
left_ele = T->val;
else
return false;
T =T->right;
}
return true;
}

};

/**
* 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 isValidBST(TreeNode* root) {
double upper = INT_MAX+1.0;
double lower = INT_MIN-1.0;
return check(root, upper,lower);

}
bool check(TreeNode* T, double upper, double lower){
if(!T)
return true;
else if(T->val>=upper||T->val<=lower)
return false;
return check(T->left,T->val,lower)&check(T->right,upper,T->val);

}

};