![Java [Leetcode 102]Binary Tree Level Order Traversal_小戴BOTAOY演示博客](/style/defalut/img/logo-bold.png){kind=logo}
题目描述：
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},文章来源地址:https://www.yii666.com/article/756106.html
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
解题思路：网址:yii666.com
运用广度优先搜索方法，同http://www.cnblogs.com/zihaowang/p/5149745.html类似。
代码如下：文章地址https://www.yii666.com/article/756106.html
/**
* 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<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> result = new LinkedList<List<Integer>>();
Queue<TreeNode> queue = new LinkedList<TreeNode>();
if(root == null)
return result;
queue.offer(root);
while(!queue.isEmpty()){
int m = queue.size();
List<Integer> list = new LinkedList<Integer>();
for(int i = 0; i < m; i++){
if(queue.peek().left != null)
queue.offer(queue.peek().left);
if(queue.peek().right != null)
queue.offer(queue.peek().right);
list.add(queue.poll().val);
}
result.add(new LinkedList<Integer>(list));
}
return result;
}
}