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.
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); } }
原文：http://www.cnblogs.com/yixianyixian/p/3716972.html