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

/**
 * 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  //第二种方法
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