113. Path Sum II

//树中可能有正数负数，所以无法剪支，开始剪支了结果导致元素出现负的时候出错/** * 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:    void get_path(vector<vector<int>> &v,vector<int> &temp,TreeNode* root,int a)    {        if(root->val==a&&!root->left&&!root->right)        {            temp.push_back(root->val);            v.push_back(temp);            temp.pop_back();            return;        }        temp.push_back(root->val);        if(root->left) get_path(v,temp,root->left,a-root->val);        if(root->right) get_path(v,temp,root->right,a-root->val);        temp.pop_back();    }    vector<vector<int>> pathSum(TreeNode* root, int sum) {        if(!root) return {};        vector<vector<int>> v;        vector<int> temp;        get_path(v,temp,root,sum);        return v;    }};