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：这是一道DFS题，当到达叶子节点时，判断从根节点到叶子节点的和是否等于sum，是的话就将结果存入，不是则回退到父节点的右节点。
class Solution {
private:
vector<vector<int> > res;
vector<int> perRes;
public:
void pathSumHelper(TreeNode* root, int pos, int sum, int target) {
if (root == NULL) {
return;
}
perRes.resize(pos+1);
perRes[pos] = root->val;
if (root->left == NULL && root->right == NULL && sum + root->val == target) {
res.push_back(perRes);
return;
}
if (root->left != NULL) {
pathSumHelper(root->left, pos+1, sum+root->val,target);
}
if (root->right != NULL) {
pathSumHelper(root->right, pos+1, sum+root->val, target);
}
}
vector<vector<int> > pathSum(TreeNode *root, int sum) {
res.clear();
if (root == NULL) {
return res;
}
perRes.clear();
pathSumHelper(root,0,0,sum);
return res;
}
};