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