leetcode102 - Binary Tree Level Order Traversal - medium

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

 

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

 

典型的BFS分层模板题

注意点是每个level的size要提前get出来，不能放在loop头里，因为queue的size是在随着pop/push变的，就错了
 
 
实现：
 1 class Solution {
 2 public:
 3     vector<vector<int>> levelOrder(TreeNode* root) {
 4         
 5         vector<vector<int>> res;
 6         if (!root) return res;
 7         
 8         queue<TreeNode*> q;
 9         q.push(root);
10         
11         while (!q.empty()){
12             vector<int> level;
13             int n = q.size();
14             for (int i=0; i<n; i++){
15                 TreeNode* cur = q.front();
16                 q.pop();
17                 level.push_back(cur->val);
18                 if (cur->left) q.push(cur->left);
19                 if (cur->right) q.push(cur->right);
20             }
21             res.push_back(level);
22         }
23         
24         return res;
25         
26     }
27 };

 

 
posted @ 2020-08-04 11:19  little_veggie  阅读(82)  评论(0)    收藏  举报