LeetCode 103: Binary Tree Zigzag Level Order Traversal

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).
For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its zigzag level order traversal as:

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        // Similar to levelorder traversal
        // However, we adopt reverse row elements every second time
        List<List<Integer>> result = new LinkedList<>();
        
        if (root == null)
            return result;

        List<Integer> row = new LinkedList<>();
        Queue<TreeNode> queue = new LinkedList<>();
        
        boolean bReverse = false; // If true, reverse the row elements
        int count = 0;
        int curMode = 1; // Number of nodes of current level
        int nxtMode = 0; // Number of nodes of next level     
        
        queue.add(root);
        
        while (!queue.isEmpty())
        {
            TreeNode node = queue.remove();
            
            row.add(node.val);
            count++;
            
            // Verify if current node is the last node of the level or not
            if (count%curMode == 0)
            {
                if (bReverse)
                {
                    reverse(row);
                    bReverse = false;
                }
                else
                    bReverse = true;
                    
                result.add(row);
                row = new LinkedList<>();
                count = 0;
            }
            
            if (node.left != null)
            {
                queue.add(node.left);
                nxtMode++; // Count the number of next level node
            }
                
            if (node.right != null)
            {
                queue.add(node.right);
                nxtMode++;
            }
            
            // When reached the end of the level, assign next mode to current mode.
            // Next mode then reset to 0.            
            if (count == 0)
            {
                curMode = nxtMode;
                nxtMode = 0;
            }
        }
        
        return result;
    }
    
    private void reverse(List<Integer> row)
    {
        int n = row.size();
        
        for (int i = 0; i < n/2; i++)
        {
            int tmp = row.get(i);
            row.set(i, row.get(n-i-1));
            row.set(n-i-1, tmp);
        }
    }
}