Leetcode 102 Binary Tree Level Order Traversal 二叉树+BFS

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<vector<int>> levelOrder(TreeNode* root) {
13         vector<vector<int>> v;
14         if(!root) return v;
15         vector<int> t;
16         
17         t.push_back(root->val);
18         v.push_back(t);
19         root->val = 1;
20         
21         queue<TreeNode*> q;
22         q.push(root);
23         
24         while(!q.empty()){
25             TreeNode* now = q.front();
26             q.pop();
27             if(!now) continue;
28             
29             q.push(now->left);
30             q.push(now->right);
31             if(now->val < v.size()){
32                 if(now->left) v[now->val].push_back(now->left->val);
33                 if(now->right) v[now->val].push_back(now->right->val);
34             }
35             else{
36                 vector<int> t;        
37                 if(now->left) t.push_back(now->left->val);
38                 if(now->right) t.push_back(now->right->val);
39                 if(t.size() != 0)v.push_back(t);
40             }
41             if(now->left) now->left->val = now->val + 1;
42             if(now->right)now->right->val = now->val + 1;
43         }
44         //reverse(v.begin(),v.end());
45         return v;
46     }
47 };