LeetCode----Path Sum

Path Sum Total Accepted: 9765 Total Submissions: 32502 My Submissions
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

题目链接

这是我的代码：

主要的注意点是要考虑 root = NULL的情况

 

 1 /**

 2  * Definition for binary tree

 3  * struct TreeNode {

 4  *     int val;

 5  *     TreeNode *left;

 6  *     TreeNode *right;

 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 8  * };

 9  */

10 class Solution {

11 public:

12     bool hasPathSum(TreeNode *root, int sum) {

13         if (root == NULL) return false;

14         if (root->left == NULL and root->right == NULL)

15             return root->val == sum ? true : false;

16         if (root->left != NULL and hasPathSum(root->left, sum-(root->val)))

17             return true;

18         if (root->right != NULL and hasPathSum(root->right, sum-(root->val)))

19             return true;

20         return false;

21     }

22 };

 

这是我的Python代码：

 1 class Solution:

 2     # @param root, a tree node

 3     # @param sum, an integer

 4     # @return a boolean

 5     def hasPathSum(self, root, sum):

 6         if root is None:

 7             return False

 8         if root.left is None and root.right is None:

 9             return True if root.val == sum else False

10         if root.left != None and self.hasPathSum(root.left, sum-(root.val)):

11             return True

12         if root.right != None and self.hasPathSum(root.right, sum-(root.val)):

13             return True

14         return False

 

Python的速度大约是C++的1/5到1/4