NC8 二叉树中和为某一值的路径(二)（C++）- 中等、树

题目链接：https://www.nowcoder.com/practice/b736e784e3e34731af99065031301bca?tpId=196&rp=1&ru=%2Fexam%2Foj&qru=%2Fexam%2Foj&sourceUrl=%2Fexam%2Foj%3Fpage%3D2%26tab%3D%25E7%25AE%2597%25E6%25B3%2595%25E7%25AF%2587%26topicId%3D196&difficulty=&judgeStatus=&tags=&title=8&gioEnter=menu
题目如下：

class Solution {
public:
    vector> FindPath(TreeNode* root,int expectNumber) {
        if(!root) return res;
        sum+=root->val;
        path.push_back(root->val);
        backtracking(root,expectNumber,sum);
        
        return res;
    }
    void backtracking(TreeNode* root,int target,int sum){
        //if(sum>=target) { //这里错误：终止条件为欸到达叶子节点，而不是这个
        if(root->left==nullptr&&root->right==nullptr){
            if(sum==target) res.push_back(path);
            return;
        }
        
        if(root->left){
            sum+=root->left->val;
            path.push_back(root->left->val);
            backtracking(root->left, target, sum);
            path.pop_back();
            sum-=root->left->val;
        }
        
        if(root->right){
            sum+=root->right->val;
            path.push_back(root->right->val);
            backtracking(root->right, target, sum);
            path.pop_back();
            sum-=root->right->val;
        }
        
    }
private:
    vector> res;
    vector path;
    int sum=0;
};