题目描述

dfs+回溯，path存储当前方案

C++ 代码

class Solution {
public:
    vector<vector<int>> res;
    int target;
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<int> path;
        target = sum;
        dfs(root, 0, path);
        return res;
    }

    void dfs(TreeNode* root, int s, vector<int>& path)
    {
        if (!root) return;
        path.push_back(root->val);
        s += root->val;
        if (s == target && !root->left && !root->right) res.push_back(path);
        dfs(root->left, s, path);
        dfs(root->right, s, path);
        path.pop_back();
    }
};