【Validate Binary Search Tree】cpp

题目：

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.

代码：

/**
 * 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) {
            if (!root) return true;
            stack<TreeNode *> sta;
            vector<int> traversal;
            TreeNode *p = root;
            while ( !sta.empty() || p )
            {
                if (p){
                    sta.push(p);
                    p = p->left;
                }
                else{
                    p = sta.top();
                    sta.pop();
                    if ( !traversal.empty() && traversal[traversal.size()-1]>=p->val ) return false;
                    traversal.push_back(p->val);
                    p = p->right;
                }
            }
            return true;
    }
};

tips：

中序遍历BST；并保留每次访问的元素；如果中序遍历的前一个元素不小于后一个元素，则返回false；全部遍历过后无问题，则返回true。

题目中有一个test case是：[-2147483648]，即INT_MIN。由于有这个非常极端的case存在，之前那种设一个pre = INT_MIN的办法就行不通了，暂时老老实实用一个vector来代替。

==========================================

一开始刷这道题总想用递归，后来看的之前的代码，BST这种判断有效的，用中序遍历可能更好一些。

/**
 * 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* curr = root;
            vector<int> vals;
            while ( !sta.empty() || curr )
            {
                if ( curr )
                {
                    sta.push(curr);
                    curr = curr->left;
                }
                else
                {
                    curr = sta.top();
                    vals.push_back(curr->val);
                    sta.pop();
                    if ( vals.size()>1 && vals[vals.size()-1]<=vals[vals.size()-2] ) return false;
                    curr = curr->right;
                }
            }
            return true;
    }
};