Leetcode: Symmetric Tree

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3
But the following is not:
    1
   / \
  2   2
   \   \
   3    3
Note:
Bonus points if you could solve it both recursively and iteratively.

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public boolean isSymmetric(TreeNode root) {
12         if (root == null) return true;
13         return check(root.left, root.right);
14     }
15     
16     public boolean check(TreeNode node1, TreeNode node2) {
17         if (node1 == null && node2 == null) return true;
18         else if (node1 == null && node2 != null) return false;
19         else if (node1 != null && node2 == null) return false;
20         else if (node1.val != node2.val) return false;
21         return check(node1.left, node2.right) && check(node1.right, node2.left);
22     }
23 }

public boolean isSymmetric(TreeNode root) {
    if(root == null)
        return true;
    if(root.left == null && root.right == null)
        return true;
    if(root.left == null || root.right == null)
        return false;
    LinkedList<TreeNode> q1 = new LinkedList<TreeNode>();
    LinkedList<TreeNode> q2 = new LinkedList<TreeNode>();
    q1.add(root.left);
    q2.add(root.right);
    while(!q1.isEmpty() && !q2.isEmpty())
    {
        TreeNode n1 = q1.poll();
        TreeNode n2 = q2.poll();
        
        if(n1.val != n2.val)
            return false;
        if(n1.left == null && n2.right != null || n1.left != null && n2.right == null)
            return false;
        if(n1.right == null && n2.left != null || n1.right != null && n2.left == null)
            return false;
        if(n1.left != null && n2.right != null)
        {
            q1.add(n1.left);
            q2.add(n2.right);
        }
        if(n1.right != null && n2.left != null)
        {
            q1.add(n1.right);
            q2.add(n2.left);
        }            
    }
    return true;
}