LeetCoder - 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) {

        if(root==null) {

            return true;

        }

        HashMap<Integer, TreeNode> lst = new HashMap<Integer, TreeNode>();

        lst.put(0, root);

        int length = 1;

        while(lst.size()!=0) {

            HashMap<Integer, TreeNode> tmp = new HashMap<Integer, TreeNode>();

            for(Integer key : lst.keySet()) {

                TreeNode node = lst.get(key);

                if(node.left!=null) {

                    tmp.put(key.intValue()*2, node.left);

                }

                if(node.right!=null) {

                    tmp.put(key.intValue()*2+1, node.right);

                }

            }

            length *= 2;

            for(Integer key : tmp.keySet()) {

                TreeNode t1 = tmp.get(key);

                TreeNode t2 = tmp.get(length-1-key.intValue());

                if((t2==null) || t1.val!=t2.val) {

                    return false;

                }

            }

            lst = tmp;

        }

        return true;

    }

}


=======
/**
 * 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 helper(root.left, root.right);
    }
   
    private boolean helper(TreeNode left, TreeNode right) {
        if (left == null && right == null) {
            return true;
        }
        if (left == null || right == null || left.val != right.val) {
            return false;
        }
        return helper(left.left, right.right) && helper(left.right, right.left);
    }
}