二叉树的序列化和反序列化

• 题目描述
  请实现两个函数，分别用来序列化和反序列化二叉树

二叉树的序列化是指：把一棵二叉树按照某种遍历方式的结果以某种格式保存为字符串，从而使得内存中建立起来的二叉树可以持久保存。序列化可以基于先序、中序、后序、层序的二叉树遍历方式来进行修改，序列化的结果是一个字符串，序列化时通过 某种符号表示空节点（#），以 ！ 表示一个结点值的结束（value!）。

二叉树的反序列化是指：根据某种遍历顺序得到的序列化字符串结果str，重构二叉树。

例如，我们可以把一个只有根节点为1的二叉树序列化为"1,"，然后通过自己的函数来解析回这个二叉树;

#include<iostream>
#include<vector>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<unordered_map>
#include <bitset>

using namespace std;
struct TreeNode
{
    int val;
    struct TreeNode* left;
    struct TreeNode* right;
    TreeNode(int x): val(x),left(nullptr),right(nullptr){};
};


class Solution {
public:
    string sHelper(TreeNode *node)
    {
        if (node == NULL)
            return "#";
        return to_string(node->val) + "," +
                sHelper(node->left) + "," +
                sHelper(node->right);
    }

    char* Serialize(TreeNode *root)
    {
        string s = sHelper(root);
        cout<<"s:"<<s<<endl;
        char *ret = new char[s.length() + 1];
        strcpy(ret, const_cast<char*>(s.c_str())); //c_str()返回一个临时的指针常量，不能对其操作;
        cout<<"*ret:"<<*ret<<"    "<<"ret:"<<ret<<endl;
        return ret;
    }

    TreeNode *dHelper(stringstream &ss)
    {
        string str;
        getline(ss, str, ','); //从输入流中读取到输出流;
        cout<<"str:  "<<str<<endl;
        if (str == "#")
            return NULL;
        else
        {
            TreeNode *node = new TreeNode(stoi(str)); //stoi将n为字符串转化为10进制的;
            node->left = dHelper(ss);
            node->right = dHelper(ss);
            return node;
        }
    }

    TreeNode* Deserialize(char *str) {
        stringstream ss(str);
        cout<<"ss:"<<ss.str()<<endl;
        return dHelper(ss);
    }
};

int main() {
        TreeNode* treeNode1 = new TreeNode(1);
        TreeNode* treeNode2 = new TreeNode(2);
        TreeNode* treeNode3 = new TreeNode(3);
        TreeNode* treeNode4 = new TreeNode(4);
        TreeNode* treeNode5 = new TreeNode(5);
        TreeNode* treeNode6 = new TreeNode(6);

        treeNode1->left = treeNode2;
        treeNode1->right = treeNode3;

        treeNode2->left = treeNode4;
        treeNode3->left = treeNode5;
        treeNode3->right = treeNode6;
        Solution sol;
        char* str = sol.Serialize(treeNode1);

        TreeNode* treeNode = sol.Deserialize(str);
    }

• 构建的二叉树为:
  
• 输出结果：中根遍历