描述
给定一个二叉树和一个目标和，找到所有从根节点到叶子节点路径总和等于给定目标和的路径。

叶子节点是指没有子节点的节点。

在线评测地址：领扣题库官网

样例1
输入: root = {5,4,8,11,#,13,4,7,2,#,#,5,1}, sum = 22
              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1
输出: [[5,4,11,2],[5,8,4,5]]
解释:
两条路径之和为 22：
5 + 4 + 11 + 2 = 22
5 + 8 + 4 + 5 = 22

样例2
输入: root = {10,6,7,5,2,1,8,#,9}, sum = 18
              10
             /  \
            6    7
          /  \   / \
          5  2   1  8
           \ 
            9 
输出: [[10,6,2],[10,7,1]]
解释:
两条路径之和为 18：
10 + 6 + 2 = 18
10 + 7 + 1 = 18

当访问的节点是叶子节点的时候，新建一个列表，插入到result中，然后返回result。 分别遍历左右子树的节点，然后将他们分别插入到叶子节点之前。

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param root: a binary tree
     * @param sum: the sum
     * @return: the scheme
     */
    vector<vector<int>> pathSum(TreeNode * root, int sum) {
        // Write your code here.
        vector<int> path;
        vector<vector<int>> Spath;
        int weight = 0;
        findpath(Spath,root,sum,path,weight);
        return Spath;
    }
    void findpath(vector<vector<int>> &Spath,TreeNode* root,int sum,vector<int> &path,int weight)
    {
        
        if(root == NULL){
            return;
        }
        weight = weight + root->val;
        path.push_back(root->val);
        if(weight == sum && root->left == NULL && root->right == NULL)
        {
            Spath.push_back(path);
            weight = weight - root->val;
            path.pop_back();
            return;
        }
        findpath(Spath,root->left,sum,path,weight);
        findpath(Spath,root->right,sum,path,weight);
        path.pop_back(); 
    }
};

更多题解参考：九章官网solution