## [LeetCode] Path Sum II

### Path Sum II

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]
]```

```/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> pathSum(TreeNode* root, int sum) {
vector<vector<int>> result;
vector<int> resultItem;
helper(result, resultItem, root, sum);
return result;
}

void helper(vector<vector<int>>& result, vector<int> resultItem, TreeNode* root, int leftSum){
if(root==NULL){
return;
}
if(root->left==NULL && root->right==NULL){  //root为叶子节点
if(leftSum==root->val){
resultItem.push_back(root->val);
result.push_back(resultItem);
}
}else{
resultItem.push_back(root->val);
leftSum = leftSum - root->val;
if(root->left!=NULL){
helper(result, resultItem, root->left, leftSum);
}
if(root->right!=NULL){
helper(result, resultItem, root->right, leftSum);
}
}
}
};```

```/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> pathSum(TreeNode* root, int sum) {
vector<vector<int>> result;
vector<int> item;
helper(root, sum, item, result);
return result;
}

void helper(TreeNode* root, int left, vector<int>& item, vector<vector<int>>& result){
if(root==NULL){
return;
}
item.push_back(root->val);
if(root->left==NULL && root->right==NULL && left == root->val){
result.push_back(item);
}
helper(root->left, left - root->val, item, result);
helper(root->right, left - root->val, item, result);
item.pop_back();
}
};```