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.

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 boolean isSymmetric(TreeNode root) {
      boolean isSys = true;
		Deque<TreeNode> nodes = new LinkedList<TreeNode>();
		if(root != null){
			nodes.addFirst(root);
			nodes.addLast(root);
			while(nodes.peek() != null){
				Deque<TreeNode> nextLevel = new LinkedList<TreeNode>();
				while(nodes.peek() != null){
					TreeNode first = nodes.pollFirst();
					TreeNode last = nodes.pollLast();
					if(first.val != last.val){
						isSys = false;
						break;
					}
					else{
						if((first.left != null && last.right == null) ||
							(first.left == null && last.right != null)){
							 isSys = false;
							 break;
						}
						if(first.left != null && last.right != null){
							nextLevel.addFirst(first.left);
							nextLevel.addLast(last.right);
						}
						if((first.right != null && last.left == null) ||
							(first.right == null && last.left != null)){
							 isSys = false;
							 break;
						}
						if(first.right != null && last.left != null){
							nextLevel.addFirst(first.right);
							nextLevel.addLast(last.left);
						}						
					}					
				}
				if(!isSys)
					break;
				else
					nodes = nextLevel;
			}			
		}
		return isSys;
    }
}