Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
For example:Given the below binary tree and
sum = 22,
5
/
4 8
/ /
11 13 4
/ /
7 2 5 1
return
[
[5,4,11,2],
[5,8,4,5]
]
思路：简单的DFS。要注意的一点是，所有的路径必须到达叶子结点才算数。叶子结点的定义是没有孩子的节点。假如root节点没有孩子节点，则root节点就算是叶子结点。但当root节点有一个左孩子节点却没有右孩子节点时，它就不能算是叶子节点了。
1 class Solution {
2 public:
3 void help(vector<vector<int> >& res, TreeNode* root, vector<int>& path, int target)
4 {
5 TreeNode *left = root->left, *right = root->right;
6 path.push_back(root->val);
7 if (!left && !right && target == root->val)
8 res.push_back(path);
9 if (left)
10 help(res, left, path, target - root->val);
11 if (right)
12 help(res, right, path, target - root->val);
13 path.pop_back();
14 }
15 vector<vector<int>> pathSum(TreeNode* root, int sum) {
16 vector<vector<int> > res;
17 if (!root) return res;
18 vector<int> path;
19 help(res, root, path, sum);
20 return res;
21 }
22 };