Binary Tree Level Order Traversal

Link: http://oj.leetcode.com/problems/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.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / 
 2   3
    /
   4
    
     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if (root == null)
            return result;
        // the algo is based on bfs, here I need a queue data structure
        ArrayList<TreeNode> queue = new ArrayList<TreeNode>();
        queue.add(root);
        ArrayList<Integer> root_value = new ArrayList<Integer>();
        // this is a special case, the value of the root should
        // be added first.
        root_value.add(root.val);
        result.add(root_value);//
        while (!queue.isEmpty()) {
            // all the nodes of the next level
            ArrayList<TreeNode> next_level = new ArrayList<TreeNode>();
            for (TreeNode t : queue) {
                // iterate through the queue
                ArrayList<TreeNode> temp_treenode = getChildren(t);
                // append all of the next level node to next_level
                next_level.addAll(temp_treenode);
            }
            // clear the queue
            queue = new ArrayList<>();
            // special case, if there's no next level nodes
            if (next_level.size() != 0) {
                queue.addAll(next_level);
                result.add(getValue(next_level));
            }
        }
        return result;
    }

    public ArrayList<Integer> getValue(ArrayList<TreeNode> list) {
        ArrayList<Integer> result = new ArrayList<Integer>();
        for (TreeNode t : list) {
            result.add(t.val);
        }
        return result;
    }

    public ArrayList<TreeNode> getChildren(TreeNode root) {
        ArrayList<TreeNode> result = new ArrayList<TreeNode>();
        if (root.left != null)
            result.add(root.left);
        if (root.right != null)
            result.add(root.right);
        return result;
    }
}

开始很纠结Queue的操作。。后来干脆遍历queue，遍历完清空queue，然后再把下一层的node再加入queue