Leetcode: Symmetric Tree

标签：leetcode   二叉树   

题目：

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.

思路分析：

题目要求分别用递归和迭代的方法进行。递归这里不说了，直接看代码。迭代算法可以使用C++标准库中的deque数据结构，允许从前面和后面操作元素。

递归算法：

class Solution
{
private:
    bool checkSymmetric(TreeNode *left, TreeNode *right)
    {
    	if (left == NULL && right == NULL)
    	{
    		return true;
    	}
    	else if (left && right && left->val == right->val)
    	{
    		return checkSymmetric(left->left, right->right) && checkSymmetric(left->right, right->left);
    	}
    	else
    	{
    		return false;
    	}
    }
public:
    bool isSymmetric(TreeNode *root)
    {
    	if (root == NULL)
    	{
    		return true;
    	}
    	else
    	{
    		return checkSymmetric(root->left, root->right);
    	}
    }
};

迭代算法（使用deque结构）：

class Solution
{
public:
    bool isSymmetric(TreeNode *root)
    {
    	if (!root)
    	{
    		return true;
    	}
    	//如果左右子树都为NULL
    	if (!root->left && !root->right)
    	{
    		return true;
    	}
    	//左右子树一个为NULL一个不为NULL
    	if ((root->left && !root->right) || (!root->left && root->right))
    	{
    		return false;
    	}
    	//左右子树都不为NULL
    	TreeNode *leftNode, *rightNode;
    	deque<TreeNode*> nodeDeque;
		nodeDeque.push_front(root->left);
		nodeDeque.push_back(root->right);
		while (!nodeDeque.empty())
		{
			leftNode = nodeDeque.front();
			rightNode = nodeDeque.back();
			nodeDeque.pop_front();
			nodeDeque.pop_back();

			if (leftNode->val != rightNode->val)
			{
				return false;
			}

			
			if ((leftNode->left && !rightNode->right) || (!leftNode->left && rightNode->right))
			{
				return false;
			}
			if (leftNode->left)
			{
				nodeDeque.push_front(leftNode->left);
				nodeDeque.push_back(rightNode->right);
			}
			
			if ((leftNode->right && !rightNode->left) || (!leftNode->right && rightNode->left))
			{
				return false;
			}
			if (leftNode->right)
			{
				nodeDeque.push_front(leftNode->right);
				nodeDeque.push_back(rightNode->left);
			}
		}
		return true;
	}
};

网上有的说采用中序遍历，判断遍历的结果对称就OK了。实际上这是不行的。

class Solution
{
private:
    void inOrder(TreeNode *node, vector<int> &result)
    {
    	if (node)
    	{
    	    inOrder(node->left, result);
    		result.push_back(node->val);
    		inOrder(node->right, result);
    	}
    }
public:
    bool isSymmetric(TreeNode *root)
    {
        if (root == NULL)
        {
            return true;
        }
    	vector<int> result;
    	inOrder(root, result);
    	vector<int>::size_type size = result.size();
    	bool flag = true;
    	for (int i = 0, j = size - 1; i < j; i++, j--)
    	{
    		if (result[i] != result[j])
    		{
    			flag = false;
    			break;
    		}
    	}
    	return flag;
    }
};

Leetcode: Symmetric Tree

标签：leetcode   二叉树   

原文地址：http://blog.csdn.net/theonegis/article/details/44199933