标签：binary tree level order traversa
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.
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}".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。
/**
* 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;
}
}
leetcode_102_Binary Tree Level Order Traversal
标签：binary tree level order traversa
原文地址：http://blog.csdn.net/mnmlist/article/details/44490975
