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 binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
struct TreeNode
{
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
#if 1 //第一种方法
class Solution {
public:
    bool isValidBST(TreeNode *root) {
		return CheckBST(root,INT_MIN,INT_MAX);
    }
private:
	bool CheckBST(TreeNode *root,int min,int max)
	{
		if(root == NULL) return true;
		return min < root->val && root->val < max && CheckBST(root->left,min,root->val) && CheckBST(root->right,root->val,max);
	}
};
#endif // 1

#if 0  //另外一种方法
class Solution {
public:
    bool isValidBST(TreeNode *root) {
		dfs(root);
		for(int i = 1; i < result.size(); i++)
		{
			if(result[i-1] >= result[i]) return false;
		}
		return true;
    }
private:
	std::vector<int> result;
	void dfs(TreeNode *root)
	{
		if(root != NULL)
		{
			dfs(root->left);
			result.push_back(root->val);
			dfs(root->right);
		}
	}
};
#endif // 1