C/C++面试题—序列化二叉树

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。

题目思路

在剑指offer中是采用的流的形式进行编写，这里将二叉树序列化为字符串，反序列化的时候将字符串反序列化为二叉树。序列化的时候用了 C++中现成的string数据结构，就不用担心空间容量的问题了。

题目代码

#include <iostream>
#include <string>
#include <fstream>
#include <vector>
#include <queue>
#include "TreeNode.h"
using namespace std;
/*

题目描述
请实现2个函数，分别用来序列化和反序列化二叉树。

*/
class SolutionSerialize
{
public:
    char* Serialize(TreeNode *root) {
        if (root == nullptr)
            return nullptr;
        string *result = new string();
        SerializeCore(root, *result);
        return (char*)(*result).data();
    }
    void SerializeCore(TreeNode *root, string &result) {
        if (root == nullptr)
            result.append("$,");
        else
        {
            result.append(to_string(root->val).data()).append(",");
            SerializeCore(root->left, result);
            SerializeCore(root->right, result);
        }
    }
    TreeNode* Deserialize(char *str) {
        if (str == nullptr)
            return nullptr;
        TreeNode* root = nullptr;
        DeserializeCore(&str, &root);
        return root;
    }
    void DeserializeCore(char** str,TreeNode** root)
    {
        char* begin = *str;
        char* sperator = strchr(begin, ',');
        if (sperator == nullptr)
            return;

        int len = sperator - begin;
        char val[20] = { 0 };
        strncpy(val, begin, len);
        *str = sperator + 1;//str指针往后移动

        if (val[0] == '$')
            *root = nullptr;
        else
        {
            *root = new TreeNode(atoi(val));
            DeserializeCore(str, &(*root)->left);
            DeserializeCore(str, &(*root)->right);
        }
    }

    //分行打印
    void PrintFromTopToBottom2(TreeNode* root) {
        //vector<int> result;
        if (root == nullptr)
            return;// result;
        queue<TreeNode*> data;
        data.push(root);
        int toBePrinted = 1;
        int nextLevel = 0;
        while (!data.empty())
        {
            TreeNode* node = data.front();
            data.pop();
            toBePrinted--;
            //result.push_back(node->val);
            cout << node->val << " ";
            if (node->left != nullptr)
            {
                data.push(node->left);
                nextLevel++;
            }
            if (node->right != nullptr)
            {
                data.push(node->right);
                nextLevel++;
            }
            if (toBePrinted == 0)
            {
                toBePrinted = nextLevel;
                nextLevel = 0;
                cout << endl;
            }
        }
        return;// result;
    }
};

int main(int argc, char *argv[])
{
    /*
        5

    2       8
1     3   6     9

    */

    SolutionSerialize solution;
    TreeNode t1(5);
    TreeNode t2(2); t1.left = &t2;
    TreeNode t3(8); t1.right = &t3;
    TreeNode t4(1); t2.left = &t4;
    TreeNode t5(3); t2.right = &t5;
    TreeNode t6(6); t3.left = &t6;
    TreeNode t7(9); t3.right = &t7;
    char* result = solution.Serialize(&t1);
    cout << result << endl;
    TreeNode *pRoot = solution.Deserialize(result);
    solution.PrintFromTopToBottom2(pRoot);
    return 0;
}

测试验证