Java for LeetCode 102 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]
]

解题思路：

前序遍历(BFS)，使用一个Queue存储每层元素即可，JAVA实现如下：

    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> list = new ArrayList<List<Integer>>();
		if (root == null)
			return list;
		Queue<TreeNode> queue = new LinkedList<TreeNode>();
		queue.add(root);
		while (queue.size() != 0) {
			List<Integer> alist = new ArrayList<Integer>();
			for (TreeNode child : queue)
				alist.add(child.val);
			list.add(new ArrayList<Integer>(alist));
			Queue<TreeNode> queue2=queue;
			queue=new LinkedList<TreeNode>();
			for(TreeNode child:queue2){
				if (child.left != null)
					queue.add(child.left);
				if (child.right != null)
					queue.add(child.right);
			}
		}
		return list;
    }