297. Serialize and Deserialize Binary Tree

最新推荐文章于 2021-06-21 12:28:59 发布

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最新推荐文章于 2021-06-21 12:28:59 发布

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分类专栏： leetcode 文章标签： bfs binary tree

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本文介绍了一种使用广度优先搜索方法来序列化和反序列化二叉树的方法。序列化将二叉树转换为字符串，而反序列化则将字符串还原为原来的二叉树结构。该方法利用队列来处理节点，确保正确地遍历每一层。

Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer,

or transmitted across a network connection link to be reconstructed later in the same or another computer environment.

Design an algorithm to serialize and deserialize a binary tree. There is no restriction on how your serialization/deserialization algorithm should work.

You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure.

For example, you may serialize the following tree

as "[1,2,3,null,null,4,5]", just the same as how LeetCode OJ serializes a binary tree. You do not necessarily need to follow this format,

so please be creative and come up with different approaches yourself.

Note: Do not use class member/global/static variables to store states. Your serialize and deserialize algorithms should be stateless.

Credits:
Special thanks to @Louis1992 for adding this problem and creating all test cases.

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二叉树的序列化和反序列化。这里都用了广度优先的方式进行，用队列实现。实现没什么难点，注意空指针就行了。

这里遇到了坑，在序列化的时候，想着一个节点用完了，后面就没有用处了，反正后面返回的是字符串，所以每个节点的内存都释放掉了。

结果通过不了，内存出错了。可能测试的时候会出现一颗二叉树用几次的情况吧。

代码：

class Codec {
public:
	// Encodes a tree to a single string.
	string serialize(TreeNode* root) 
	{
		string res;
		if(!root) return "null";
		queue<TreeNode*>q;
		q.push(root);
		while(!q.empty())
		{
			TreeNode* tmp = q.front();
			q.pop();
			if(!tmp)	
			{
				res += "null,";
			}
			else
			{
				res += to_string(tmp->val) + ",";
				q.push(tmp->left);
				q.push(tmp->right);
			}
		}
		return res;
	}

	// Decodes your encoded data to tree.
	TreeNode* deserialize(string data) 
	{
		int pos = 0, n = data.size();
		data += ",";
		TreeNode* root = build_node(data, pos);
		if(!root) return NULL;
		queue<TreeNode*>q;
		q.push(root);
		while(!q.empty())
		{
			TreeNode* tmp = q.front();
			q.pop();
			if(pos >= n) break;
			tmp->left = build_node(data, pos);
			if(tmp->left) q.push(tmp->left);
			if(pos >= n) break;
			tmp->right = build_node(data, pos);
			if(tmp->right) q.push(tmp->right);
		}
		return root;
	}
private:
	TreeNode* build_node(const string& s, int& pos)
	{
		int end = s.find(",", pos);
		string tmp = s.substr(pos, end-pos);
		pos = end + 1;
		if(tmp == "null")
		{
			return NULL; 
		}
		else
		{
			TreeNode *node = new TreeNode(stoi(tmp));
			return node;
		}
	}
};