C++ How to declare as atribute, a pointer to my own type, using templates -

this problem: have class "nodo" (to make bsts) has array of pointers same type attribute, , have son class "nodoavl" (to make avl trees) have inherit atribute, pointers must point same type, is, point "nodoavl". there way this? maybe code, can explain better:

#include<iostream> using namespace std;  template <class t> class nodo  {  protected:   t d;   "my_own_type" *h[2] = {null};  // here need pointers own type                                   // nodo has 2 pointers nodo                                  // , nodoavl has 2 pointers nodoavl  public:   nodo(t dato = t()) { d = dato; }  };  template <class t> class nodoavl : public nodo<t>  {  protected:   int alt = 0;  public:   nodoavl (t dato = t()): nodo<t>(dato) {}   int altu () {return alt;}  }; 

i gratefull help.

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--- update ---

i have read crtp (thanks some programmer dude) , although didn't understand utility, found solve problem. decided make "nodo" big-father class adding 1 label in template "tn" (type of node), array of pointers declared tn*. created 2 son class: nodobst , nodoavl, each 1 inherit "nodo" labeled respective type of node in template. again code explain better idea.

#include<iostream> using namespace std;  template <class t, class tn> class nodo  {  protected:   t d;   tn *h[2] = {null};  public:   nodo(t dato = t()) { d = dato; }  };  template <class t> class nodobst : public nodo<t,nodobst<t>>  {  protected:  public:   nodobst (t dato = t()): nodo<t,nodobst<t>>(dato) {}  };  template <class t> class nodoavl : public nodo<t, nodoavl<t>>  {  protected:   int alt = 0;  public:   nodoavl (t dato = t()): nodo<t,nodoavl<t>>(dato) {}   int altu () {return alt;}  };  int main ()  {  nodoavl<int> * narval = new nodoavl<int>(10);  cout << narval->altu();  } 

it has worked, want know if it's practice or i'm making future problems. want add more methods father class works array of pointers , of them same in kind of node (ej: update pointers, dereference them, kill them, etc). if had solve problem? lot again!

you can obvious way:

nodo<t>* h[2]; 

but can using class name alone:

nodo* h[2];