原题链接在这里：

题目：

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]]

题解：

是一道BFS，利用queue，先进先出，并利用curCount来判断是否走完了当前一层。当curCount = 0 时表示当前一层走完，要把当前list加入res里，再刷新list，问题是不要忘记同时刷新curCount和nextCount.

Time Complexity: O(n). Space: O(n).

AC Java:

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List
     
      
       > levelOrder(TreeNode root) {        List
       
        
         > res = new ArrayList
         
          
           >(); if(root == null){ return res; } LinkedList
           
             que = new LinkedList
            
             (); que.add(root); int curCount = 1; int nextCount = 0; List
             
               item = new ArrayList
              
               (); while(!que.isEmpty()){ TreeNode tn = que.poll(); item.add(tn.val); curCount--; if(tn.left != null){ que.add(tn.left); nextCount++; } if(tn.right != null){ que.add(tn.right); nextCount++; } if(curCount == 0){ res.add(new ArrayList
               
                (item)); item = new ArrayList
                
                 (); curCount = nextCount; nextCount = 0; } } return res; }}
                
               
              
             
            
           
          
         
        
       
      
     

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