Java binary tree sequence traversal (traversal layer by layer, different from zigzag printing)

Title description

Given a binary tree, return the result of the traversal of the binary tree sequence, (traversing layer by layer from left to right),
for example:
the given binary tree is {3,9,20,#,#,15,7},

The result of this binary tree sequence traversal is
[
[3],
[9,20],
[15,7]
]

Example 1

enter

{1,2}

return value

[[1],[2]]

Example 2

enter

{1,2,3,4,#,#,5}

return value

[[1],[2,3],[4,5]]

import java.util.*;

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 * }
 */

public class Solution {
    /**
     * 
     * @param root TreeNode类 
     * @return int整型ArrayList<ArrayList<>>
     */
    public ArrayList<ArrayList<Integer>> levelOrder (TreeNode root) {
       
        ArrayList<ArrayList<Integer>> reult = new ArrayList<ArrayList<Integer>>();
        
        if(root == null){
            return reult;
        }
        
        ArrayList<Integer> list = null;
        int queueSize = 0;
        LinkedList<TreeNode> queue = new LinkedList();
        queue.offer(root);
        
        while(!queue.isEmpty()){            
            
            list = new ArrayList();
            
            //先存取之前队列中的长度 这样就保证了之前的长度代表二叉树 同一层中节点的个数
            queueSize = queue.size();
            
            //队列中的长度到0 则同一层的出队完成了
            while(queueSize-- > 0){
                TreeNode temp = queue.poll();
                list.add(temp.val);

                if(temp.left!=null){
                    queue.offer(temp.left);
                }

                if(temp.right!=null){
                    queue.offer(temp.right);
                }
            }
            
            reult.add(list);
        }
        
        return reult;
      
    }
}