Question

我的程序正在调用一个成员函数，该函数使用通过引用传递的指针执行插入到二进制搜索树中（以便更改指针的值）。但是，当我尝试将第二个节点插入树时程序崩溃。我假设这与坏指针有关。

可以更改此分配的唯一部分是成员函数的实现，而不是声明。

任何帮助都将不胜感激。

 //Header file
using namespace std;

struct PersonRec
{
    char name[20];
    int bribe;
    PersonRec* leftLink;
    PersonRec* rightLink;
};


class CTree
{

private:
    PersonRec *tree;
    bool IsEmpty();
    void AddItem( PersonRec*&, PersonRec*);
    void DisplayTree(PersonRec*);

public:
    CTree();
    //~CTree();
    void Add();
    void View();

};

//Implementation file

#include <iostream>
#include <string>

using namespace std;

#include "ctree.h"

CTree::CTree()
{
    tree = NULL;
}

//PersonList::~MyTree()
//{
//
//}


bool CTree::IsEmpty()
{
    if(tree == NULL) 
    {
        return true;
    }
    else
    {
        return false;
    }
}

void CTree::Add()
{
    PersonRec* newPerson = new PersonRec();
    cout << "Enter the person's name: ";
    std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
    cin.getline(newPerson->name, 20);
    cout << "Enter the person's contribution: ";
    cin >> newPerson->bribe;


    newPerson->leftLink = NULL;
    newPerson->rightLink = NULL;

    AddItem(tree, newPerson);
}

void CTree::View()
{
    if (IsEmpty())
    {
        cout<<"The list is empy";
    }
    else
    {
        DisplayTree(tree);

    }

};

void CTree::AddItem( PersonRec*& ptr, PersonRec* newPer )
{
        if (tree == NULL)
        {
            ptr = newPer;
        }
        else if ( newPer->bribe < ptr->bribe)
            AddItem(ptr->leftLink, newPer); 
        else
            AddItem(ptr->rightLink, newPer); 
}
void CTree::DisplayTree(PersonRec* ptr)
{
    if (ptr == NULL)
                    return;
    DisplayTree(ptr->rightLink);
    cout<<ptr->name<<" "<<"$"<<ptr->bribe <<endl;
    DisplayTree(ptr->leftLink); 
}


//Driver file
#include <iostream>

using namespace std;
#include <cstdlib>
#include "ctree.h"

int displayMenu (void);
void processChoice(int, CTree&);

int main (void)
{
int num;
CTree ct;
do 
{
num = displayMenu();
if (num != 3)
processChoice(num, ct);
} while (num != 3);
return 0;
}

int displayMenu (void)
{
int choice;
cout << "\nMenu\n";
cout << "==============================\n\n";
cout << "1. Add student to waiting list\n";
cout << "2. View waiting list\n";
cout << "3. Exit program\n\n";
cout << "Please enter choice: ";
cin >> choice;
return choice;
}

void processChoice(int choice, CTree& myTree)
{
   switch (choice)
   {
      case 1: myTree.Add (); break;
      case 2: myTree.View (); break;
   } 
}

Answer 1

在CTree::AddItem中，您的情况有误：

    if (tree == NULL)
    {
        ptr = newPer;
    }

应该是

    if (ptr == NULL)
    {
        ptr = newPer;
    }

当您致电AddItem(ptr->rightLink, newPer);时，newPer是新节点，但当前根的后代中的任何一个都可以是NULL，因此您需要检查{{1}并且替换，而不是新节点。

通过引用传递指针（插入二进制搜索树）

1 个答案: