101. Symmetric Tree

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101. Symmetric Tree

题目

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

Note:
Bonus points if you could solve it both recursively and iteratively. 

解析

• 递归和迭代实现

// 101. Symmetric Tree
class Solution_101 {
public:

	// 递归调用，同时判断左子树的左节点与右子树的右节点，以及左子树的右节点与右子树的左节点。一旦这两个节点不相等，就返回false。 
	bool isSymmetricHelp(TreeNode *left, TreeNode* right)
	{
		if (!left&&!right)
		{
			return true;
		}
		if (!left&&right ||left&&!right)
		{
			return false;
		}
		if (left->val!=right->val)
		{
			return false;
		}

		return isSymmetricHelp(left->left, right->right) && isSymmetricHelp(left->right, right->left);
	}

	bool isSymmetric(TreeNode *root) {

		if (!root)
		{
			return true;
		}
		return isSymmetricHelp(root->left, root->right);
	}

	//非递归实现
	//需要：对每一层成对送入队列，出队列比较
	bool isSymmetric1(TreeNode* root) {
		if (!root)
		{
			return true;
		}
		queue<TreeNode*> que;

		que.push(root->left);
		que.push(root->right);

		while (!que.empty())
		{
			int size = que.size();
			while (size)
			{
				TreeNode* left = que.front();
				que.pop();
				TreeNode* right = que.front(); //取出成对的元素
				que.pop();
				size -= 2;
				if (!left&&!right)
				{
					continue;
				}
				if (!left&&right)
				{
					return false;
				}
				if (left&&!right)
				{
					return false;
				}

				if (left->val!=right->val)
				{
					return false;
				}
				que.push(left->left);
				que.push(right->right);
				que.push(left->right);
				que.push(right->left);
			}
		}
		return true;
	}
};

101. Symmetric Tree