1 class Solution {
2 /**
3 * @param root: The root of binary tree.
4 * @return: Level order a list of lists of integer
5 */
6 struct node{
7 TreeNode* tn;
8 int cnt;
9 node(TreeNode* tn, int cnt){
10 this->tn = tn;
11 this->cnt = cnt;
12 }
13 };
14
15 public:
16 void dfs(vector<vector<int>> &ans, queue<node*> &que){
17 if(que.empty()) return ;
18 node* now = que.front();
19 que.pop();
20
21 if(now->tn->left) que.push(new node(now->tn->left, now->cnt + 1));
22 if(now->tn->right) que.push(new node(now->tn->right, now->cnt + 1));
23
24 if((int)ans.size() - 1 < now->cnt){
25 vector<int> v(1,now->tn->val);
26 ans.push_back(v);
27
28 } else {
29 ans[now->cnt].push_back(now->tn->val);
30 }
31
32 dfs(ans, que);
33
34 }
35 vector<vector<int>> levelOrder(TreeNode *root) {
36 // write your code here
37 vector<vector<int>> ans;
38 if(root == NULL) return ans;
39 queue<node*> que;
40 que.push(new node(root,0));
41 dfs(ans, que);
42 return ans;
43 }
44 };
