【LeetCode剑指offer34】二叉树中和为某一值的路径（dfs回溯）

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>>ans;
    vector<int>temp;
    int sum = 0;
    vector<vector<int>> pathSum(TreeNode* root, int target) {
        dfs(root, target);
        return ans;
    }
    void dfs(TreeNode* root, int target){
        if(root == NULL){
            return;
        }
        temp.push_back(root->val);
        sum += root->val;
        //到根节点了，并且sum满足条件
        if(root->left == NULL && root->right == NULL && sum == target){
            ans.push_back(temp);
        }
        dfs(root->left, target);
        dfs(root->right, target);
        //当左右孩子都为空的节点时，就把temp里最后的节点弹出，dfs弹出节点的兄弟节点
        temp.pop_back();
        sum -= root->val;
    }
};