如何从代码中删除类并使其正常工作

Question

所以为了学校，我编写了代码并使用了类，但老师说我不应该使用类
这是一种编辑我当前代码的方法吗
这是一个用类创建的二进制搜索树，我需要删除它。
关于要做什么、删除或更改什么的信息会有所帮助，您无需编辑我的代码。

#include <iostream>
#include <cstdlib>
using namespace std;
class BST
{
private:
struct tree_node
{
tree_node* left;
tree_node* right;
int data;
};

public:
BST()
{
root = NULL;
parent = NULL;
}
tree_node* root;
tree_node* parent;
bool isEmpty() const { return root==NULL; }
void print_inOrder();
void inOrder(tree_node*);
void print_preOrder();
void preOrder(tree_node*);
void print_postOrder();
void postOrder(tree_node*);
void search(tree_node*,int,tree_node*);
void insert(int);
void create(int);
void remove(int);
};
void BTS::search(tree_node* root,int k,tree_node* parent)
{
if (root == nullptr)
{
cout << "Element not found"<<endl;
return;
}
if (root->data == k)
{
if (parent == nullptr)
cout << "element " << k << " found "; //jeigu ieskomas skaicius medzio virsune
else cout << "element "<<k<<" found "<<endl;
return;
}
if (k < root->data)
return searcch(root->left,k, root); // rekursija jeigu reiksme mazesne uz sakny einame i kaire
return search(root->right,k,root);// rekursija jeigu reiksme mazesne uz sakny einame i desine
}
void BTS::create(int d)
{
for(int i=0;i<d;i++)
{
tree_node* t = new tree_node;
tree_node* parent;
cout<<"enter element"<<endl;
cin>>t->data;
t->left = NULL;
t->right = NULL;
parent = NULL;
if(isEmpty()) root = t;
else
{
tree_node* current;
current = root;
while(current)
{
parent = current;
if(t->data > current->data) current = current->right;
else current = current->left;
}
if(t->data < parent->data)
parent->left = t;
else
parent->right = t;
}
}
}
void BTS::remove(int d)
{
bool found = false;
if(isEmpty())
{
cout<<" tree is empty "<<endl;
return;
}
tree_node* curr;
tree_node* parent;
curr = root;
while(curr != NULL)
{
if(curr->data == d)
{
found = true;
break;
}
else
{
parent = curr;
if(d>curr->data) curr = curr->right;
else curr = curr->left;
}
}
if(!found)
{
cout<<" element not found"<<endl;
return;
}

if((curr->left == NULL && curr->right != NULL)|| (curr->left != NULL
&& curr->right == NULL))
{
if(curr->left == NULL && curr->right != NULL)
{
if(parent->left == curr)
{
parent->left = curr->right;
delete curr;
}
else
{
parent->right = curr->right;
delete curr;
}
}
else
{
if(parent->left == curr)
{
parent->left = curr->left;
delete curr;
}
else
{
parent->right = curr->left;
delete curr;
}
}
return;
}
if( curr->left == NULL && curr->right == NULL)
{
if(parent->left == curr) parent->left = NULL;
else parent->right = NULL;
delete curr;
return;
}
if (curr->left != NULL && curr->right != NULL)
{
tree_node* chkr;
chkr = curr->right;
if((chkr->left == NULL) && (chkr->right == NULL))
{
curr = chkr;
delete chkr;
curr->right = NULL;
}
else
{
if((curr->right)->left != NULL)
{
tree_node* lcurr;
tree_node* lcurrp;
lcurrp = curr->right;
lcurr = (curr->right)->left;
while(lcurr->left != NULL)
{
lcurrp = lcurr;
lcurr = lcurr->left;
}
curr->data = lcurr->data;
delete lcurr;
lcurrp->left = NULL;
}
else
{
tree_node* tmp;
tmp = curr->right;
curr->data = tmp->data;
curr->right = tmp->right;
delete tmp;
}
}
return;
}
}
void BTS::print_inOrder()
{
inOrder(root);
}
void BTS::inOrder(tree_node* p)// naudojant rekursija atspausdinu medi vidniu apejimu
{
if(p != NULL)
{
if(p->left) inOrder(p->left);
cout<<" "<<p->data<<" ";
if(p->right) inOrder(p->right);
}
else return;
}

void BTS::print_preOrder()
{
preOrder(root);
}
void BTS::preOrder(tree_node* p)// naudojant rekursija atspausdinu medi tiesioginu apejimu
{
if(p != NULL)
{
cout<<" "<<p->data<<" ";
if(p->left) preOrder(p->left);
if(p->right) preOrder(p->right);
}
else return;
}
void BTS::print_postOrder()
{
postOrder(root);
}
void BTS::postOrder(tree_node* p) // naudojant rekursija atspausdinu medi atvikstiniu apejiu
{
if(p != NULL)
{
if(p->left) postOrder(p->left);
if(p->right) postOrder(p->right);
cout<<" "<<p->data<<" ";
}
else return;
}
void BTS::insert(int d)
{
tree_node* t = new tree_node;
tree_node* parent;
t->data = d;
t->left = NULL;
t->right = NULL;
parent = NULL;
if(isEmpty()) root = t;
else
{
tree_node* current;
current = root;
while(current)
{
parent = current;
if(t->data > current->data) current = current->right;
else current = current->left;
}
if(t->data < parent->data)
parent->left = t;
else
parent->right = t;
}
}
int main()
{
BTS b;
int punktas=0,c,d;
cout<<endl<<endl;
cout<<" MENU "<<endl;
cout<<" ----------------------------- "<<endl;
cout<<" 1) sukurti "<<endl;
cout<<" 2) tiesioginis elementu spausdinimas "<<endl;
cout<<" 3) vidinis elementu spausdinimas "<<endl;
cout<<" 4) atvirkstinis elementu spausdinimas "<<endl;
cout<<" 5) iterpti nauja elementa "<<endl;
cout<<" 6) reiksmes trinimas "<<endl;
cout<<" 7) reiksmes paieska "<<endl;
cout<<" 0) Iseiti "<<endl;
cout<<" Iveskite norima punkta"<<endl;
cin>>punktas;
int tikrinimas = 1;
while(punktas!=0)
{
if(punktas==1)
{cout<<"Kiek reiksmiu bus sarase?"<<endl;
cin>>c;
b.sukurti(c); //panaudojam klases sukurti funckija
tikrinimas = 0;
}
else if(punktas==2 && tikrinimas == 0)
{
cout<<" tiesioginis atspausdinimas "<<endl;
cout<<" -------------------"<<endl;
b.print_tiesioginis();// panaudojam klases tiesioginis atspausdinias funckija
}
else if(punktas==3 && tikrinimas == 0)
{
cout<<" vidinis atspausdinimas "<<endl;
cout<<" -------------------"<<endl;
b.print_vidinis();// panaudoja klases funkcija
}
else if(punktas==4 && tikrinimas == 0)
{
cout<<" atvirkstinis atspausdinimas "<<endl;
cout<<" --------------------"<<endl;
b.print_atvirkstinis();// panaudoja klases funkcija
}
else if(punktas==5 && tikrinimas == 0)
{cout<<"elemento iterpimas"<<endl;
cout<<" --------------------"<<endl;
cin>>c;
b.iterpti(c);// panaudoja klases funkcija
}
else if(punktas==6 && tikrinimas == 0)
{
cout<<" iveskite elementa kuri norite istrinti "<<endl;
cin>>d;
b.salinti(d);// panaudoja klases funkcija
}
else if(punktas==7 && tikrinimas == 0)
{
cout<<"iveskite reiksme kuria norite surasti"<<endl;
cin>>c;
b.paieska(b.root,c,b.tevas);
}
else if(punktas<8){cout<<"sukurkyte medi norit atlikti punktus"<<endl;}
cout<<"iveskite kita punkta"<<endl;
cin>>punktas;
}
}

Answer 1

您并没有在这里使用太多类。

您已经创建了一个名为b的paieskosMedis，并将一些应用程序逻辑放入其数据成员和成员函数中。

因此，只需将它们移动到自由函数和全局变量中即可。这可能是你老师想要的。

例如：

class Foo
{
public:
Foo() { m = 3; }
void bar() { m = 4; }
private:
int m;
};
int main()
{
Foo f;
f.bar();
}

变为：

int m;
void bar() { m = 4; }
int main()
{
m = 3;
bar();
}

而且，嘿，你可能已经解决了下一周的任务。

注意，tree_node仍然是一个类，因为struct引入了类，但你的老师可能没有意识到这一点，并将它们视为一些单独的"结构"。所以我们会忽略这一点。

Answer 2

谢谢你的帮助，如果我惹恼了你，我很抱歉。通过将成员函数移动到自由函数来修复它。