原题链接：39.组合总和

思路：
依旧是遍历整棵树，此时终止条件变为总和sum大于给定的target值 或者等于target 就进行回溯

全代码：

class Solution {
public:vector<vector<int>> result;vector<int> path;void backtracking(vector<int>& candidates,int target, int sum, int startIndex){if(sum > target) return ; //如果综合大于目标数，则进行回溯if(sum == target){//如果等于目标数，则将组合存储进容器中，并进行回溯result.push_back(path);return;}for(int i = startIndex; i < candidates.size(); i++){sum += candidates[i];  //经过一个结点则值相加path.push_back(candidates[i]);//将值pushback进容器内backtracking(candidates,target,sum,i);//递归遍历sum -= candidates[i];   //回溯path.pop_back();//回溯}}vector<vector<int>> combinationSum(vector<int>& candidates, int target) {result.clear();path.clear();backtracking(candidates,target,0,0);return result;}
};