原文链接：http://www.cnblogs.com/riasky/p/3508668.html

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.

利用递归法，可以很轻松解决的问题。

增加一个adaptor接口函数就可以了，

搞清楚对称的判断方法：每一个左子树与右子树的节点相等，然后是左子树的右子树和右子树的左子树相等， 有点拗口，不过画图就很好理解了。

 

bool isSymmetric(TreeNode *root) 
	{
		if (!root) return true;
		return symmetric(root->left, root->right);
	}

	bool symmetric(TreeNode *t1, TreeNode *t2)
	{
		if (!t1 && !t2) return true;
		if (!t1 && t2 || t1 && !t2) return false;
		if (t1->val != t2->val) return false;

		return symmetric(t1->left, t2->right) && symmetric(t1->right, t2->left);
	}

 

 

转载于:https://www.cnblogs.com/riasky/p/3508668.html