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 }

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