297. Serialize and Deserialize Binary Tree

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

    1   / /  2   3     / /    4   5as "[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;		}	}};