Populating Next Right Pointers in Each Node - Perfect Binary Tree

class Solution {
    
    public Node connect(Node root) {
        
        //Get level order traversal of the BT
        List<List<Node>> levels = levelOrder(root);

        //Iterate through each level
        for(int i=0; i<levels.size(); i++){

            //Get the level i
            List<Node> level = levels.get(i);

            //For each Node, set the next pointer to its next
            for(int j=0; j<level.size()-1; j++){
            level.get(j).next = level.get(j+1);
            }

            //We don't really need to update the next for the last node since it would already be null
            //level.get(level.size()-1).next = null;

        }
        return root;
    }
    
    //Get the Level Order Traversal on the BT
    public List<List<Node>> levelOrder(Node root) {
        
        //If the BST is null
        if(root == null) return new ArrayList<List<Node>>();
        
        //We will make the Queue type as TreeNode because we need to access the left and right children later
        Queue<Node> q = new LinkedList<>();

        //Initialize the queue with the root node and NULL
        //The idea is to add NULL after every level
        //While removing elements from the queue, if we encounter a NULL, that means that we have covered a certain level
        //So we can append the current list to the answer and continue
        q.add(root);
        q.add(null);
        
        //To store final answer
        List<List<Node>> res = new ArrayList<>();

        //To store each level
        List<Node> temp = new ArrayList<>();
        
        //If there is one node and NULL, size must be at least 1
        while(q.size() > 1){

            //Remove the element from the queue
            Node f = q.remove();

            //If the element is NULL, that means, we have completed a level. 
            if(f == null){
                //Add NULL to end of the queue so mark the next level because at this moment, we must have completed adding all childer of the current level
                q.add(null);
                //Add current level to answer
                res.add(temp);
                //Reset array so store next level
                temp = new ArrayList<>();
            }else{
                //Else, add the current element to current level
                temp.add(f);
                //Add childer to the queue
                if(f.left != null) q.add(f.left);
                if(f.right != null) q.add(f.right);
            }
        }
        
        //Add final level
        res.add(temp);
        
        return res;
        
    }
    
}