/**
 * 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) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> res;
    vector<vector<int>> levelOrder(TreeNode* root) {      
        if(root==NULL)
        {
            return res;
        }
        DFS(1,root);
        return res;
        /*
        queue<TreeNode*> q;
        vector<vector<int>> res;
        if(root==NULL)
        {
            return res;
        }
        q.push(root);
        while(!q.empty())
        {
            //当前存在于队列中的都是该层的
            int curlength=q.size();
            vector<int> temp;
            for(int i=0;i<curlength;i++)
            {
                 TreeNode* node=q.front();
                 q.pop();
                 temp.push_back(node->val);
                 if(node->left)
                     q.push(node->left);
                 if(node->right)
                     q.push(node->right);
            }
            res.push_back(temp);
        }
        return res;
        */
    }
    //思路大致为 从root开始，进入第一层房间，然后root安顿好后，把孩子们往下一层安顿
    void DFS(int index,TreeNode* root)
    {
        if(root==NULL)
        {
            return;
        }
        //还没有容纳当前层的房子，就新建一个      
        if(res.size()<index)
        {
            vector<int> temp;
            res.push_back(temp);
        }
        res[index-1].push_back(root->val);
        if(root->left)
        {
            DFS(index+1,root->left);
        }
        if(root->right)
        {
            DFS(index+1,root->right);
        }
    }
};