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>>res;
    vector<int>path;
    void dfs(TreeNode* root, int number)
    {
        if(root==nullptr)
        {
            return;
        }
        path.emplace_back(root->val);
        number = number - root->val;

        if(root->left==nullptr&&root->right==nullptr&&number==0)
        {
            res.emplace_back(path);
        }

        dfs(root->left, number);
        dfs(root->right, number);
        path.pop_back();
    }
    vector<vector<int> > FindPath(TreeNode* root, int target) {
        // write code here
        dfs(root, target);
        return res;
    }
};