//0045二叉树中和为某一数的路径
//错误代码：会重复访问路径，问题在子节点的判断
//这样的访问没得救
class Solution {
public:
    vector<int>buff;
    vector<vector<int>>ans;

    void Dfs(TreeNode *root,int cur, int target)
    {
        //每个叶子结点都有空的左右节点，所以不能这样判断叶子节点：这是叶节点的左右儿子
        //所以会两次访问
        if (root == NULL) {
            if (cur == target) {
                ans.push_back(buff);
            }
            return;
        }
        buff.push_back(root -> val);
        Dfs(root -> left, cur + root -> val, target);
        Dfs(root -> right, cur + root -> val, target);
        buff.pop_back();
    }

    vector<vector<int>> findPath(TreeNode* root, int sum) {
        buff.clear();
        ans.clear();
        if (root == NULL) return ans;
        Dfs(root, 0, sum);
        ans.erase(unique(ans.begin(), ans.end()), ans.end());
        return ans;
    }
};
//正确的
class Solution {
public:
    vector<int>buff;
    vector<vector<int>>ans;

    void Dfs(TreeNode *root,int cur)
    {
        if (root == NULL) return;
        cur -= root -> val;
        buff.push_back(root -> val);
        if (!root -> left && !root -> right && !cur) ans.push_back(buff);
        Dfs(root -> left, cur);
        Dfs(root -> right, cur);
        buff.pop_back();
    }

    vector<vector<int>> findPath(TreeNode* root, int sum) {
        buff.clear();
        ans.clear();
        Dfs(root, sum);
        return ans;
    }
};