LeetCode: Sum Root to Leaf Numbers

少数次过

 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 dfs(TreeNode *root, int &ret, int tmp) {
13         if (!root) return;
14         if (!root->left && !root->right) ret += tmp*10+root->val;
15         else {
16             if (root->left) dfs(root->left, ret, tmp*10+root->val);
17             if (root->right) dfs(root->right, ret, tmp*10+root->val);
18         }
19     }
20     int sumNumbers(TreeNode *root) {
21         // Start typing your C/C++ solution below
22         // DO NOT write int main() function
23         if (!root) return 0;
24         int ret = 0;
25         dfs(root, ret, 0);
26         return ret;
27     }
28 };

C#

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public int val;
 5  *     public TreeNode left;
 6  *     public TreeNode right;
 7  *     public TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public int SumNumbers(TreeNode root) {
12         if (root == null) return 0;
13         int ans = 0;
14         dfs(root, ref ans, 0);
15         return ans;
16     }
17     public void dfs(TreeNode root, ref int ans, int tmp) {
18         if (root == null) return;
19         if (root.left == null && root.right == null) ans += tmp * 10 + root.val;
20         else {
21             if (root.left != null) dfs(root.left, ref ans, tmp * 10 + root.val);
22             if (root.right != null) dfs(root.right, ref ans, tmp * 10 + root.val);
23         }
24     }
25 }

View Code

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原文地址：https://www.cnblogs.com/yingzhongwen/p/3034089.html