[Leetcode][Tree][Sum Root to Leaf Numbers]

非常简单的一个题，和path sum非常类似。

 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     void sum(TreeNode *root, int now, int &res) {
13         if (root == NULL) {
14             return;
15         }
16         now = now * 10 + root->val;
17         if (root->left == NULL && root->right == NULL) {
18             res += now;
19         }
20         sum(root->left, now, res);
21         sum(root->right, now, res);
22     }
23     int sumNumbers(TreeNode *root) {
24         int res = 0;
25         sum(root, 0, res);
26         return res;
27     }
28 };

但是看完题解之后，发现题解中的解法还是比我的算法更优秀！赶紧学习！

 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     int sum(TreeNode *root, int now) {
13         if (root == NULL) {
14             return 0;
15         }
16         now = now * 10 + root->val;
17         if (root->left == NULL && root->right == NULL) {
18             return now;
19         }
20         return sum(root->left, now) + sum(root->right, now);
21     }
22     int sumNumbers(TreeNode *root) {
23         return sum(root, 0);
24     }
25 };

更少的参数，更短的代码！

 

posted @ 2014-07-05 17:03  poemqiong  阅读(126)  评论(0)  编辑  收藏  举报