Leetcode Symmetric Tree 递归和非递归解法

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.

和same tree题目差不多一样的解法。

1 recursively 递归法：

//2014-2-15 update
	bool isSymmetric(TreeNode *root) 
	{
		if (!root) return true;
		return isSymmetric(root->left, root->right);
	}
	bool isSymmetric(TreeNode *lt, TreeNode *rt)
	{
		if (!lt && !rt) return true;
		if (lt && !rt || !lt && rt || lt->val != rt->val) return false;
		return isSymmetric(lt->left, rt->right) &&isSymmetric(lt->right, rt->left);
	}

2 iteratively 非递归：

使用先序遍历，原来树的先序遍历的非递归写法也不简单。

//2014-2-15 update
	bool isSymmetric(TreeNode *root) 
	{
		if (!root || !root->left && !root->right) return true;
		TreeNode *t1 = root->left, *t2 = root->right;
		if (t1&&!t2 || !t1&&t2 || t1->val!=t2->val) return false;

		stack<TreeNode *> s1, s2;
		s1.push(t1), s2.push(t2);
		bool flag = false;
		while (!s1.empty() && !s2.empty())
		{
			if (!flag && (t1->left || t2->right))
			{
				s1.push(t1), s2.push(t2);
				t1 = t1->left, t2 = t2->right;
				if (t1&&!t2 || !t1&&t2 || t1->val!=t2->val) return false;
				s1.push(t1), s2.push(t2);
			}
			else if (t1->right || t2->left)
			{
				t1 = t1->right, t2 = t2->left;
				if (t1&&!t2 || !t1&&t2 || t1->val!=t2->val) return false;
				flag = false;
			}
			else
			{
				t1 = s1.top(), t2 = s2.top();
				s1.pop(), s2.pop();
				flag = true;
			}
		}
		return s1.empty() && s2.empty();
	}