普通的BFS.
唯一小心就是：每次要把每一Level 的放在一个list里面。也就是说，一旦queue里有东西，那一定就全部是这一个level的。先把size固定一下，把queue里面的东西全部加到这个level的list里面（同时还要继续添加element 进 queue）。没跑完一圈i ~ size， 然后就add那个level list。

/*
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.

Tags:  Tree Breadth-first Search
Similar Problems: (M) Binary Tree Zigzag Level Order Traversal, (E) Binary Tree Level Order Traversal II, (E) Minimum Depth of Binary Tree

*/

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
/*
Thoughts:
Looks like BFS. put root in a queue, then put left, right in the queue.
Process the queue until it runs out.

Note: 
Queue: offer()


*/
public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> rst = new ArrayList<List<Integer>>();
        if (root == null) {
            return rst;
        }
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);

        while (!queue.isEmpty()) {
            int size = queue.size();//Becareful with fixed size
            List<Integer> list = new ArrayList<Integer>();
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                list.add(node.val);
                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
            }
            rst.add(list);
        }
        return rst;
    }
}