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; }
 * }
 * 非递归的解法，可以用两个双端队列做层序遍历，然后队列从两端弹出元素，两端元素相等说明是对称
 */
public class Solution {
    public boolean isSymmetricHelper(TreeNode r1, TreeNode r2) {
        //左右半支到叶子节点的时候 判断相等
        if (r1 == null && r2 == null) return true;
        if (r1 == null && r2 != null) return false;
        if (r1 != null && r2 == null) return false;
        if (r1.val != r2.val) return false;
        //左半支的左子树 和右半支的右子树比较相等 左半支的右子树 和右半支的左子树比较相等
        return isSymmetricHelper(r1.left, r2.right) && isSymmetricHelper(r1.right,r2.left);
    }

    public boolean isSymmetric(TreeNode root) {
        if (root==null){
            return true;
        }
        if (root.left==null||root.right==null){
            return root.left == root.right;
        }
        return isSymmetricHelper(root.left,root.right);
    }
}