Given an n-ary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

和Binary Tree的做法一样

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;

    public Node() {}

    public Node(int _val,List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/
class Solution {
    public List<List<Integer>> levelOrder(Node root) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if(root == null)
            return res;

        LinkedList<Node> curr = new LinkedList<Node>();
        LinkedList<Node> next = new LinkedList<Node>();
        List<Integer> line = new ArrayList<Integer>();
        curr.offer(root);

        while(curr.size() > 0) {
            Node temp = curr.poll();
            line.add(temp.val);

            if(temp.children != null && temp.children.size() > 0) {
                for(Node child: temp.children)
                    next.offer(child);
            }

            if(curr.size() == 0) {
                res.add(line);
                line = new ArrayList<Integer>();
                curr = next;
                next = new LinkedList<Node>();
            }
        }

        return res;
    }
}