判断二叉树是否是镜像

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.

OJ\'s Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where \'#\' signifies a path terminator where no node exists below.

Here\'s an example:

   1
  / \\
 2   3
    /
   4
    \\
     5

The above binary tree is serialized as\"{1,2,3,#,#,4,#,#,5}\".

package leetcode;

public class Symmetric_tree {
     public static void main(String[] args) {
        TreeNode t1 = new TreeNode(1);
        TreeNode t2 = new TreeNode(2);
        t1.right = t2;
        System.out.println(isSymmetric(t1));
    }
     
     public static boolean isSymmetric(TreeNode root) {
         if(root == null || (root.left == null && root.right == null)) {
             return true;
         }
         //traverse(root.right);        //这个位置我当时是想通过递归使右子树整个转置,然后判断左右子树进行相等
         if(isSame(root.left,root.right)) {    //
             return true;
         }else {
             return false;
         }
     }

    private static boolean isSame(TreeNode left, TreeNode right) {
        // TODO Auto-generated method stub
        
        if((left == null && right == null)) {
            return true;
        }else if(left != null && right != null) {
            if(left.val == right.val)
            return isSame(left.left, right.right) && isSame(left.right, right.left); //其实不需要转置,只要判断左的右和右的左是否相等
            //return isSame(left.left, right.left) && isSame(left.right, right.right); //脑子抽了,还想转置之后判断左和左与右和右
        }
        return false;
    }

    /*private static void traverse(TreeNode root) {     //利用递归使这个树进行转置
        // TODO Auto-generated method stub
        if(root == null) {
            return;
        }else {
            if(root.left != null) {
                traverse(root.left);
            }
            if(root.right != null) {
                traverse(root.right);
            }
            TreeNode temp = root.left;
            root.left = root.right;
            root.right = temp;
        }
            
    }*/
     
}