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);
}

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