Binary Tree Level Order Traversal

Binary Tree Level Order Traversal

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]
]

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public List<List<Integer>> levelOrder(TreeNode root) {
12         List<List<Integer>> container = new ArrayList<List<Integer>>();
13         if(null == root)
14             return container;
15         //List<TreeNode> theSameLevel = new ArrayList<TreeNode>();//存放同一层节点
16         Queue<TreeNode> theSameLevel = new LinkedList<TreeNode>();
17         theSameLevel.add(root);
18         while(!theSameLevel.isEmpty()){//theSameLevel不为空
19             List<Integer> oneLevel = new ArrayList<Integer>();
20             Queue<TreeNode> temp = new LinkedList<TreeNode>();//暂存同一层的结点
21             while(!theSameLevel.isEmpty()){
22                 TreeNode cur = theSameLevel.remove();
23                 oneLevel.add(cur.val);
24                 if(null != cur.left) 
25                     temp.add(cur.left);
26                 if(null != cur.right)
27                     temp.add(cur.right);
28             }
29             theSameLevel = temp;
30             container.add(oneLevel);
31         }
32         
33         return container;
34     }
35 }