【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.

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}"

.

/**
* 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 isMir(root.left,root.right);
}
return true;
}

private boolean isMir(TreeNode left, TreeNode right) {
// TODO Auto-generated method stub
if(left==null&&right!=null)
return false;
if(left!=null&&right==null)
return false;
if(left==null&&right==null)
return true;
if(left.val!=right.val)
return false;

return isMir(left.left,right.right)&&isMir(left.right,right.left);
}
}