问题描述：
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.
分析：判断给定的一棵二叉树是不是对称的。
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */
public boolean isSymmetric(TreeNode root) { //递归 if (root == null) return true; return isMiror(root.left, root.right); } public boolean isMiror(TreeNode n1, TreeNode n2) { //判断两棵树是否成镜像 if (n1 == null && n2 == null) return true; if (n1 == null && n2 != null) return false; if (n1 != null && n2 == null) return false; if (n1.val != n2.val) return false; return isMiror(n1.left, n2.right) && isMiror(n1.right, n2.left); }
原文：http://www.cnblogs.com/mydesky2012/p/5056172.html