标签:des style http color io os ar for sp
Given a binary tree, determine if it is a valid binary search tree (BST).
Assume a BST is defined as follows:
confused what "{1,#,2,3}" means? >
read more on how binary tree is serialized on OJ.
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 //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 // 1leetcode - Validate Binary Search Tree
标签:des style http color io os ar for sp
原文地址:http://blog.csdn.net/akibatakuya/article/details/39829137