题解 | #二叉树中和为某一值的路径(二)#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
#include <vector>
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 target int整型 
     * @return int整型vector<vector<>>
     */
    vector<vector<int> > FindPath(TreeNode* root, int target) {
        // write code here
        vector<vector<int>> result;
        vector<int> path;
        dfs(result, path, root, target);
        return result;
    }
    void dfs(vector<vector<int>>& result, vector<int>& path, TreeNode *root, int target) {
        if (root == nullptr) {
            return;
        }
        cout << "root->val = " << root->val << " target = " << target;
        path.push_back(root->val);
        target -= root->val;
        if (root->left == nullptr && root->right == nullptr && target == 0) {
            result.push_back(path);
            cout << "res" << endl;
        }
        dfs(result, path, root->left, target);
        dfs(result, path, root->right, target);
        path.pop_back();
    }
};