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## Problem

Given an n-ary tree, return the level order traversal of its nodes’ values.

Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).

Example 1:

Input: root = [1,null,3,2,4,null,5,6]
Output: [[1],[3,2,4],[5,6]]


Example 2:

Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
Output: [[1],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]


## Code

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

public Node() {}

public Node(int _val) {
val = _val;
}

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<>();
if(root == null) return res;

queue.offer(root);

while(!queue.isEmpty()) {
int size = queue.size();
List<Integer> curr = new ArrayList<>();
for(int i = 0; i < size; i++) {
Node node = queue.poll();