leetcode【第十五周】路径求和 【二】

问题描述：

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]
]

问题分析：

如题所述，很容易可以想到使用深度优先搜索遍历二叉树，然后维护一个数组，用来保存从根节点到当前节点所经过的路径的节点值，当遍历到叶子节点时，对访问路径所经过的节点的值求和，若与欲求的目标值相等，则为目标路径，否则返回，尝试下一条路径。

具体实现：

/**
 * 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> pathSum(TreeNode* root, int sum) {
        vector> res;
        vector list;
        searchSum(res,root,list,0,sum);
        
    }
    void searchSum(vector> &res, TreeNode* root, vector list, int val, int sumTarget)
    {
        if(res==NULL)
        {
            if(val==sumTarget)
                res.pushback(list);
            return;
        }
        list.pushback(root->val);
        val = val+root->val;
        searchSum(res, root->left,list,val,sumTarget);
        searchSum(res, root->right,list,val,sumTarget);
    
    }
    
    
};