## 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]
]


### Analysis:

Classic DFS search similar to the Question Path Sum I, in this problem we just add a vector<int> to store the result.
Details see code below:

### Code(C++):

/**
* 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:
void ps(TreeNode* root, int sum, vector<int> path, vector<vector<int>> &res){
if (!root){
return;
}else{
path.push_back(root->val);
if (!root->left && !root->right && sum==root->val){
res.push_back(path);
path.clear();
}
ps(root->left, sum - root->val, path, res);
ps(root->right, sum - root->val, path, res);
}
}
vector<vector<int>> pathSum(TreeNode* root, int sum) {
vector<vector<int>> res;
vector<int>path;
ps(root, sum, path, res);
return res;
}
};


### Code(Python):

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
def ps(self, root, sum, path, res):
if not root:
return
else:
path.append(root.val)
if not root.left and not root.right and sum == root.val:
res.append(path)
path = []
else:
self.ps(root.left, sum - root.val, path, res)
self.ps(root.right, sum - root.val, path, res)

def pathSum(self, root, sum):
"""
:type root: TreeNode
:type sum: int
:rtype: List[List[int]]
"""
res = []
path = []
self.ps(root, sum, path, res)
return res