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],
]
 1 public class Solution {
 2     public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
 3         ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
 4         if(root==null) return res;
 5         LinkedList<TreeNode> cur = new LinkedList<TreeNode>();
 6         cur.offer(root);
 7         while(!cur.isEmpty()){
 8             LinkedList<TreeNode> next = new LinkedList<TreeNode>();
 9             ArrayList<Integer> temp = new ArrayList<Integer>();
10             for(TreeNode n:cur){
11                 temp.add(n.val);
12                 if(n.left!=null)
13                     next.offer(n.left);
14                 if(n.right!=null)
15                     next.offer(n.right);
16             }
17             cur = next;
18             res.add(temp);
19         }
20         Collections.reverse(res);
21         return res;
22     }
23 }
View Code

 

posted @ 2014-02-22 13:11  krunning  阅读(160)  评论(0)  编辑  收藏  举报