101 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.

Code: 1st Attempt:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private boolean symmetric = true;
    public boolean isSymmetric(TreeNode root) {
        if (root == null) {
            return true;
        }
        helper(root.left, root.right);
        return symmetric;
    }
    private void helper(TreeNode lroot, TreeNode rroot) {
        if (lroot == null && rroot == null) {
            return;
        }
        if ((lroot == null && rroot != null) || (lroot != null && rroot == null)) {
            symmetric = false;
            // if no return here, lroot.left, rroot.right will have NullPointerException
            return;
        }
        if (lroot.val == rroot.val) {
            helper(lroot.left, rroot.right);
            helper(lroot.right, rroot.left);
        } else {
            symmetric = false;
        }
    }
}

Updated based on w,(12.18.15)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) {
            return true;
        }
        return isSymmetric(root.left, root.right);
    }

    private boolean isSymmetric(TreeNode left, TreeNode right) {
        if (left == null && right == null) {
            return true;
        }
        if (left == null || right == null) {
            return false;
        }
        return (left.val == right.val) && isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left);
    }
}