LeetCode: Path Sum II

多数次过

 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 sum, vector<vector<int>> &ret, vector<int> &tmp) {

13         if (!root) return;

14         tmp.push_back(root->val);

15         if (sum == root->val && !root->left && !root->right) ret.push_back(tmp);

16         if (root->left) dfs(root->left, sum-root->val, ret, tmp);

17         if (root->right) dfs(root->right, sum-root->val, ret, tmp);

18         tmp.pop_back();

19     }

20     vector<vector<int> > pathSum(TreeNode *root, int sum) {

21         // Start typing your C/C++ solution below

22         // DO NOT write int main() function

23         vector<vector<int>> ret;

24         vector<int> tmp;

25         dfs(root, sum, ret, tmp);

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 List<List<int>> PathSum(TreeNode root, int sum) {

12         List<List<int>> ans = new List<List<int>>();

13         List<int> tmp = new List<int>();

14         dfs(root, sum, ref ans, ref tmp);

15         return ans;

16     }

17     public void dfs(TreeNode root, int sum, ref List<List<int>> ans, ref List<int> tmp) {

18         if (root == null) return;

19         tmp.Add(root.val);

20         if (sum == root.val && root.left == null && root.right == null) ans.Add(new List<int>(tmp.ToArray()));

21         if (root.left != null) dfs(root.left, sum - root.val, ref ans, ref tmp);

22         if (root.right != null) dfs(root.right, sum -  root.val, ref ans, ref tmp);

23         tmp.RemoveAt(tmp.Count-1);

24     }

25 }

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