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   1return[   [5,4,11,2],   [5,8,4,5]]

分析：

这道题目算是的延伸。仅仅只是要求出全部从根到树叶的路径上节点和为给定值的路径，并放到一个vector中返回。

採取相似于递归思路。每次递归都会建立一个新的vector，复制之前的vector内容，并将当前节点的值放入。假设当前节点是叶节点。而且路径上节点和满足要求。就将新的vector放入终于返回的vector中。

代码：

class Solution {public:    vector
      
       
        > res;    vector
        
         
          > pathSum(TreeNode* root, int sum) {         if(!root)            return res;        vector
          
            tmp;        pathSum1(root,sum,tmp);         return res;    }    void pathSum1(TreeNode* root, int sum,vector
           
             tmp) { if(!root) return ; vector
            
              tmp1; tmp1 = tmp; tmp1.push_back(root->val); if(!root->left&&!root->right) { if(root->val==sum) res.push_back(tmp1); return; } pathSum1(root->left,sum-root->val,tmp1);pathSum1(root->right,sum-root->val,tmp1); }};