给定一个二叉树和一个和，找到所有从根到叶路径总和等于给定总和的路径。
例如，
给定下面的二叉树和 sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
返回
[
[5,4,11,2],
[5,8,4,5]
]
详见：https://leetcode.com/problems/path-sum-ii/description/
/**
* 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>> res;
vector<int> path;
if(root==nullptr)
{
return res;
}
findPath(root,sum,path,res);
return res;
}
void findPath(TreeNode* root,int sum,vector<int> &path,vector<vector<int>> &res)
{
if(root==nullptr)
{
return;
}
path.push_back(root->val);
if(root->val==sum&&root->left==nullptr&&root->right==nullptr)
{
res.push_back(path);
path.pop_back();
}
else
{
findPath(root->left,sum-root->val,path,res);
findPath(root->right,sum-root->val,path,res);
path.pop_back();
}
}
};
