Why if statement is not entering when Null - c++

Question

I have the next code, and when raiz is NULL is not entering the if statement
I already tried to do the other way like if(raiz != NULL) and in the else statement trying to run the code :
masAlto = 1;
raiz = Nuevo;
Here is my code:
void InsertaAVL(arbolAVL *raiz, arbolAVL *Nuevo, int masAlto)
{
arbolAVL *a = raiz;
int subarbolmasAlto = FALSO;
if (raiz == NULL)
{
masAlto = 1;
raiz = Nuevo;
}
else
{
if (raiz->llave > Nuevo->llave)
{
//Insertar a la Izquierda
InsertaAVL(raiz->Izq, Nuevo, subarbolmasAlto);
if (subarbolmasAlto == VERDADerO)
{
switch(raiz->FBalance)
{
case LH : BalanceIzquierdo(raiz,masAlto);
break;
case RH : raiz->FBalance = EH;
masAlto = FALSO;
break;
case EH : raiz->FBalance = LH;
masAlto = VERDADerO;
break;
}
}
else
masAlto = FALSO;
}
else
{
InsertaAVL(raiz->Der, Nuevo, subarbolmasAlto);
if (subarbolmasAlto == VERDADerO)
{
switch(raiz->FBalance)
{
case LH : raiz->FBalance = EH;
break;
case RH : BalanceDerecho(raiz,masAlto);
break;
case EH : raiz->FBalance = RH;
masAlto = VERDADerO;
break;
}
}
else
masAlto = FALSO;
}
}
}
I expect that when raiz is NULL to enter the code inside the if statement

Answer 1

It does enter the if statement true branch, although I'd prefer if (!raiz) as the conditional as this behaves well with the NULL of C and the nullptr of C++. Effectively though the true branch is a no-op since the assignments
masAlto = 1;
raiz = Nuevo;
have no effect on the quantities passed by the caller of the function. Did you want int* masAlto &c. as the parameters?