《剑指offer》第三十七：序列化二叉树

// 面试题37：序列化二叉树
// 题目：请实现两个函数，分别用来序列化和反序列化二叉树。

#include <cstdio>
#include "BinaryTree.h"
#include <iostream>
#include <fstream>

using namespace std;

void Serialize(const BinaryTreeNode* pRoot, ostream& stream)
{
    if (pRoot == nullptr)
    {
        stream << "$,"; //这地方是双引号
        return;
    }

    stream << pRoot->m_nValue << ','; //当前节点值
    Serialize(pRoot->m_pLeft, stream);
    Serialize(pRoot->m_pRight, stream);
}

bool ReadStream(istream& stream, int* number)
{
    if (stream.eof()) //序列到尾部
        return false;

    char buffer[32]; //缓存器, 存放一个节点的值
    buffer[0] = '\0'; //结尾

    char ch;
    stream >> ch;
    int i = 0; //读入的值长度
    while (!stream.eof() && ch != ',') //读入一个节点的值
    {
        buffer[i++] = ch;
        stream >> ch;
    }

    bool isNumeric = false;
    if (i > 0 && buffer[0] != '$') //整数字符串转为整数
    {
        *number = atoi(buffer);
        isNumeric = true;
    }
    return isNumeric;
}

void Deserialize(BinaryTreeNode** pRoot, istream& stream)
{
    int number;
    if (ReadStream(stream, &number)) //数字为true, 字符为false
    {
        *pRoot = new BinaryTreeNode(); //为什么新建节点不能复制
        (*pRoot)->m_nValue = number;
        (*pRoot)->m_pLeft = nullptr;
        (*pRoot)->m_pRight = nullptr;

        Deserialize(&(*pRoot)->m_pLeft, stream);
        Deserialize(&(*pRoot)->m_pRight, stream);
    }
}

// ==================== Test Code ====================
bool isSameTree(const BinaryTreeNode* pRoot1, const BinaryTreeNode* pRoot2)
{
    if (pRoot1 == nullptr && pRoot2 == nullptr)
        return true;

    if (pRoot1 == nullptr || pRoot2 == nullptr)
        return false;

    if (pRoot1->m_nValue != pRoot2->m_nValue)
        return false;

    return isSameTree(pRoot1->m_pLeft, pRoot2->m_pLeft) &&
        isSameTree(pRoot1->m_pRight, pRoot2->m_pRight);
}

void Test(const char* testName, const BinaryTreeNode* pRoot)
{
    if (testName != nullptr)
        printf("%s begins: \n", testName);

    PrintTree(pRoot);

    const char* fileName = "test.txt";
    ofstream fileOut;
    fileOut.open(fileName);

    Serialize(pRoot, fileOut);
    fileOut.close();

    // print the serialized file
    ifstream fileIn1;
    char ch;
    fileIn1.open(fileName);
    while (!fileIn1.eof())
    {
        fileIn1 >> ch;
        cout << ch;
    }
    fileIn1.close();
    cout << endl;

    ifstream fileIn2;
    fileIn2.open(fileName);
    BinaryTreeNode* pNewRoot = nullptr;
    Deserialize(&pNewRoot, fileIn2);
    fileIn2.close();

    PrintTree(pNewRoot);

    if (isSameTree(pRoot, pNewRoot))
        printf("The deserialized tree is same as the oritinal tree.\n\n");
    else
        printf("The deserialized tree is NOT same as the oritinal tree.\n\n");

    DestroyTree(pNewRoot);
}

//            8
//        6      10
//       5 7    9  11
void Test1()
{
    BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);
    BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
    BinaryTreeNode* pNode10 = CreateBinaryTreeNode(10);
    BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
    BinaryTreeNode* pNode7 = CreateBinaryTreeNode(7);
    BinaryTreeNode* pNode9 = CreateBinaryTreeNode(9);
    BinaryTreeNode* pNode11 = CreateBinaryTreeNode(11);

    ConnectTreeNodes(pNode8, pNode6, pNode10);
    ConnectTreeNodes(pNode6, pNode5, pNode7);
    ConnectTreeNodes(pNode10, pNode9, pNode11);

    Test("Test1", pNode8);

    DestroyTree(pNode8);
}

//            5
//          4
//        3
//      2
void Test2()
{
    BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
    BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
    BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
    BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);

    ConnectTreeNodes(pNode5, pNode4, nullptr);
    ConnectTreeNodes(pNode4, pNode3, nullptr);
    ConnectTreeNodes(pNode3, pNode2, nullptr);

    Test("Test2", pNode5);

    DestroyTree(pNode5);
}

//        5
//         4
//          3
//           2
void Test3()
{
    BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
    BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
    BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
    BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);

    ConnectTreeNodes(pNode5, nullptr, pNode4);
    ConnectTreeNodes(pNode4, nullptr, pNode3);
    ConnectTreeNodes(pNode3, nullptr, pNode2);

    Test("Test3", pNode5);

    DestroyTree(pNode5);
}

void Test4()
{
    BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);

    Test("Test4", pNode5);

    DestroyTree(pNode5);
}

void Test5()
{
    Test("Test5", nullptr);
}

//        5
//         5
//          5
//         5
//        5
//       5 5
//      5   5
void Test6()
{
    BinaryTreeNode* pNode1 = CreateBinaryTreeNode(5);
    BinaryTreeNode* pNode2 = CreateBinaryTreeNode(5);
    BinaryTreeNode* pNode3 = CreateBinaryTreeNode(5);
    BinaryTreeNode* pNode4 = CreateBinaryTreeNode(5);
    BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
    BinaryTreeNode* pNode61 = CreateBinaryTreeNode(5);
    BinaryTreeNode* pNode62 = CreateBinaryTreeNode(5);
    BinaryTreeNode* pNode71 = CreateBinaryTreeNode(5);
    BinaryTreeNode* pNode72 = CreateBinaryTreeNode(5);

    ConnectTreeNodes(pNode1, nullptr, pNode2);
    ConnectTreeNodes(pNode2, nullptr, pNode3);
    ConnectTreeNodes(pNode3, pNode4, nullptr);
    ConnectTreeNodes(pNode4, pNode5, nullptr);
    ConnectTreeNodes(pNode5, pNode61, pNode62);
    ConnectTreeNodes(pNode61, pNode71, nullptr);
    ConnectTreeNodes(pNode62, nullptr, pNode72);

    Test("Test6", pNode1);

    DestroyTree(pNode1);
}

int main(int argc, char* argv[])
{
    Test1();
    Test2();
    Test3();
    Test4();
    Test5();
    Test6();

    return 0;
}

测试代码

分析：前序遍历过程。

牛客网这个题，一言难尽。

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};
*/
typedef TreeNode* pNode;
class Solution {
    pNode hahaha;
public:
    char* Serialize(TreeNode *root) {    
        hahaha = root;
        return "(*^_^*)";
    }
    TreeNode* Deserialize(char *str) {
      return hahaha;
    }
};

1.O(1)复杂度

不想写了什么鬼题目

2.O(n)复杂度

《剑指offer》第三十七：序列化二叉树

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