LeetCode - Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

 

return its bottom-up level order traversal as:

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

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 ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
12         // Start typing your Java solution below
13         // DO NOT write main() function
14         
15         ArrayList<ArrayList<Integer>> result = new  ArrayList<ArrayList<Integer>>();
16         LinkedList<TreeNode> currLevel = new LinkedList<TreeNode>();
17         
18         int level = 0;
19         if(root != null) currLevel.add(root);
20         while(!currLevel.isEmpty()){
21             LinkedList<TreeNode> nextLevel = new LinkedList<TreeNode>();
22             ArrayList<Integer> list = new ArrayList<Integer>();
23             for(TreeNode n : currLevel){
24                 list.add(n.val);
25                 if(n.left != null){
26                     nextLevel.add(n.left);
27                 }
28                 if(n.right != null){
29                     nextLevel.add(n.right);
30                 }
31             }
32             currLevel = nextLevel;
33             result.add(list);
34             level++;
35         }
36         Collections.reverse(result);
37         return result;
38     }
39 }

 

posted @ 2013-01-29 09:07  cradle  阅读(212)  评论(0)  编辑  收藏  举报