判断是否为二叉查找树的题，先给一种方法，可是过不去某个case
/**
* 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:
//这种方法当test case为[-2147483648,null,2147483647]或[-2147483648］或[2147483647]时无法通过
bool isValidBST(TreeNode* root)
return helper(root, INT_MIN, INT_MAX);
}
bool helper(TreeNode* root, int min, int max){
if(root == NULL) return true;
if(root->val <= min || root->val >= max)
return false;
return helper(root->left, min, root->val) && helper(root->right, root->val, max);
}
};
换另外一种方法（In-order tranversal）:
/**
* 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)
{
TreeNode* pre = NULL;
return helper(root, pre);
}
bool helper(TreeNode* root, TreeNode* &pre){
if(root == NULL) return true;
if(!helper(root->left, pre)) return false;
if(pre != NULL && pre->val >= root->val)
return false;
pre = root;
return helper(root->right, pre);
}
};
这题值得好好回味一下
