题解 | #牛牛的路径总和#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
#include <algorithm>
#include <cstddef>
#include <ios>
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @param targetSum int整型 
     * @return int整型vector<vector<>>
     */
    void backtravel(vector<vector<int>>& res, vector<int>& path, TreeNode* root, int& targetSum)
    {
        path.push_back(root->val);
        targetSum -= root->val;
        if(root->left == nullptr && root->right == nullptr)
        {
            if(targetSum == 0)
                res.push_back(path);
        }
        if(root->left)
        {
            backtravel(res, path, root->left, targetSum);
            path.pop_back();
            targetSum += root->left->val;
        }
        if(root->right)
        {
            backtravel(res, path, root->right, targetSum);
            path.pop_back();
            targetSum += root->right->val;
        }
        return;
    }

    vector<vector<int>> findPaths(TreeNode* root, int targetSum) {
        // write code here
        vector<vector<int>> res;
        vector<int> path;
        backtravel(res, path, root, targetSum);
        return  res;
    }
};