leetcode_102_Binary Tree Level Order Traversal

1.描述

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,#,#,15,7}

,

3
/ \
9  20
/  \
15   7

return its level order traversal as:

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

confused what 

"{1,#,2,3}"

 means? >
read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:

1
/ \
2   3
/
4
\
5

The above binary tree is serialized as 

"{1,2,3,#,#,4,#,#,5}"

.

2.思路

一般的层序遍历直接打印出结果，用队列即可，但是此次的要求尼是按层次打印结果，所以考虑到用两个队列来交替存储，遍历上一层次的同时将下一层的结点存储到另一个队列中，并在将上面一层的遍历完成后交换两个队列的值。

3.代码

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>>list=new ArrayList<List<Integer>>();//存储结果
if(root==null)
return list;
Queue<TreeNode>q1=new LinkedList<TreeNode>();//交替存储相邻两层的结点
Queue<TreeNode>q2=new LinkedList<TreeNode>();
Queue<TreeNode>temp=null;
List<Integer>subList=null;//存储一层的结点的值
q1.add(root);
TreeNode top=null;
while(!q1.isEmpty())
{
subList=new ArrayList<Integer>();
while(!q1.isEmpty())//循环遍历一层结点并将下一层结点存储到队列中
{
top=q1.peek();
q1.poll();
if(top.left!=null)
q2.add(top.left);
if(top.right!=null)
q2.add(top.right);
subList.add(top.val);
}
list.add(subList);
temp=q2;//交换两个队列的值，使q1一直指向要遍历的那一层
q2=q1;
q1=temp;
}
return list;
}
}

4.结果