标签：
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 [1,2,2,3,4,4,3] is symmetric:
1 / 2 2 / \ / 3 4 4 3
But the following [1,2,2,null,3,null,3] is not:
1 / 2 2 \ 3 3
算法分析：和same tree比较像，都可以用递归方法来解决。
public class SymmeticTree
{
public boolean isSymmetric(TreeNode root)
{
if (root == null)
{
return true;
}
return isSymmetric(root.left, root.right);
}
public boolean isSymmetric(TreeNode left, TreeNode right)
{
if (left == null && right == null)
{
return true;
}
else if ((left == null && right != null)
|| (right == null && left != null) || (left.val != right.val))
{
return false;
}
else
{
return isSymmetric(left.left, right.right)
&& isSymmetric(left.right, right.left);
}
}
}
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原文地址：http://www.cnblogs.com/masterlibin/p/5903854.html
