Leetcode-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]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

Solution:

 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>> res = new ArrayList<List<Integer>>();
13         List<Integer> oneRes = new ArrayList<Integer>();
14         if (root==null){
15             return res;
16         }
17 
18         List<TreeNode> lastLevel = new ArrayList<TreeNode>();
19         lastLevel.add(root);
20         oneRes.add(root.val);
21         res.add(oneRes);
22         List<TreeNode> curLevel;
23         while (lastLevel.size()!=0){
24             oneRes = new ArrayList<Integer>();
25             curLevel = new ArrayList<TreeNode>();
26             for (int i=0;i<lastLevel.size();i++){
27                 TreeNode curNode = lastLevel.get(i);
28                 if (curNode.left!=null){
29                     curLevel.add(curNode.left);
30                     oneRes.add(curNode.left.val);
31                 }
32                 if (curNode.right!=null){
33                     curLevel.add(curNode.right);
34                     oneRes.add(curNode.right.val);
35                 }
36             }
37             if (curLevel.size()==0){
38                 break;
39             } else {
40                 res.add(oneRes);
41                 lastLevel = curLevel;
42             }
43         }
44 
45         return res;        
46     }
47 }

 

posted @ 2014-11-29 05:07  LiBlog  阅读(116)  评论(0)  编辑  收藏  举报