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

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]]

The snippet shows n-ary level order traversal

private void levelOrdertraversal(Node node) {

    //find out their parent level if the parent level is the same then they belong to same group
    //at each level update child level, if they belong to same level then while traversal group them

        int arr[] = new int[1000];
        List<Node> list = new ArrayList<>();
        List<List<Node>> listOfNodes = new ArrayList<>();
        Queue<Node> queue = new LinkedList<>();
        queue.add(node);
        arr[node.val] = 0;

        list.add(node);
        listOfNodes.add(list);

        while(queue.isEmpty()) {
            Node currentnode = queue.poll();
            for (Node child : node.children){
                arr[currentnode.val] += 1;
                if(listOfNodes.get(arr[currentnode.val])== null) {
                    ArrayList<Node> newL = new ArrayList<>();
                    newL.add(child);
                } else {
                    List<Node> l = listOfNodes.get(arr[currentnode.val]);
                    l.add(child);
                }
                queue.add(child);
            }
        }
    }