107 Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

My solution:

class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> ans = new ArrayList<List<Integer>>();
        if(root == null) return ans;
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        q.add(root);
        while(q.isEmpty() == false){
            Object[] t = q.toArray();
            q.clear();
            List<Integer> l = new ArrayList<Integer>();
            for(int i = 0; i < t.length; i++) {
                TreeNode node = (TreeNode)t[i];
                l.add(node.val);
                if (node.left != null) {
                    q.add(node.left);
                }
                if (node.right != null) {
                    q.add(node.right);
                }
            }
            ans.add(0,l);
        }
        return ans;
    }
}

2 things.

1. Why is my solution DFS, not BFS?
2. Improvements can be made in my code. For example, using the length of the queue and poll method, we don't need to remove all the items in the queue before we append new item to the queue.
   Others' solution:

DFS solution:

public class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        List<List<Integer>> wrapList = new LinkedList<List<Integer>>();
        
        if(root == null) return wrapList;
        
        queue.offer(root);
        while(!queue.isEmpty()){
            int levelNum = queue.size();
            List<Integer> subList = new LinkedList<Integer>();
            for(int i=0; i<levelNum; i++) {
                if(queue.peek().left != null) queue.offer(queue.peek().left);
                if(queue.peek().right != null) queue.offer(queue.peek().right);
                subList.add(queue.poll().val);
            }
            wrapList.add(0, subList);
        }
        return wrapList;
    }
}
BFS solution:

public class Solution {
        public List<List<Integer>> levelOrderBottom(TreeNode root) {
            List<List<Integer>> wrapList = new LinkedList<List<Integer>>();
            levelMaker(wrapList, root, 0);
            return wrapList;
        }
        
        public void levelMaker(List<List<Integer>> list, TreeNode root, int level) {
            if(root == null) return;
            if(level >= list.size()) {
                list.add(0, new LinkedList<Integer>());
            }
            levelMaker(list, root.left, level+1);
            levelMaker(list, root.right, level+1);
            list.get(list.size()-level-1).add(root.val);
        }
    }