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//输入一棵二叉树和一个整数，打印出二叉树中节点值的和为输入整数的所有路径。从树的根节点开始往下一直到叶节点所经过的节点形成一条路径。
/**
 * 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) {}
 * };
 */
#include <iostream>
#include <vector>
using namespace std;

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) {}
};

vector<vector<int>> ans;
vector<int> cur_path;

vector<vector<int>> pathSum(TreeNode* root, int target) {
    if (root == nullptr) {
        return ans;
    }
    cur_path.push_back(root->val);
    if (root->val == target && root->left == nullptr && root->right == nullptr) { //是叶子结点且和为target 
        ans.push_back(cur_path);//满足条件 入ans
    }
    pathSum(root->left, target - root->val);
    pathSum(root->right, target - root->val);
    if (cur_path.size() > 0) { //当前节点出栈
        cur_path.pop_back(); 
    }
    return ans;
}

int main() {
    TreeNode *left1 = new TreeNode(1);
    TreeNode *right = new TreeNode(3);
    TreeNode *right1 = new TreeNode(5);
    TreeNode *left = new TreeNode(2, left1, right1);
    TreeNode *root = new TreeNode(2, left, right);
    pathSum(root, 5);
    for (int i = 0; i < ans.size(); i++) {
        for(int j = 0; j < ans[i].size(); j++) {
            cout << ans[i][j] << " ";
        }
        cout << endl;
    }
    return 0;
}