Java Solution, using level order traversal using 2 queues Start with putting 1 in the primary queue, 1. poll from primary queue, check if you need to swap the child i.e level % k == 0. 2. add the child indexes in secondary queuue. repeat step 1 and 2 until primary queue is empty. once empty means level is done, go to next level. poll all the indexes from secondary and add into primary. do the inOrderTraversal of the tree and add the list to a list

public static List> swapNodes(List> indexes, List queries) { // Write your code here Queue primary = new LinkedList<>(); Queue secondary = new LinkedList<>();

    List<List<Integer>> res = new ArrayList<>();
    for (int k : queries) {
        primary.add(1);
        int level = 1;
        while (!primary.isEmpty()) {
            while (!primary.isEmpty()) {
                int p = primary.poll();
                List<Integer> child = indexes.get(p - 1);
                if (level % k == 0) { // swap left and right child
                    child.add(child.get(0));
                    child.remove(0);
                }
                child.stream().filter(ch -> ch != -1).forEach(secondary::add); //add the non null child to secondary queue representing next level
            }
            while (!secondary.isEmpty())
                primary.add(secondary.poll());
            level++;
        }
        List<Integer> traversal = new ArrayList<>();
        inOrderTraversal(1, indexes, traversal);
        res.add(traversal);
    }

    return res;
}

private static void inOrderTraversal(int root, List<List<Integer>> indexes, List<Integer> res) {
    int left = indexes.get(root - 1).get(0);
    if (left != -1)
        inOrderTraversal(left, indexes, res);
    res.add(root);
    int right = indexes.get(root - 1).get(1);
    if (right != -1)
        inOrderTraversal(right, indexes, res);
}