Doubts about pointer pointers and function returns

Question

Hello, folks. I’d like to know why the function below would need a pointer to pointer, in case a **root, for the line /1/ could modify the *root, while the line /2/ can modify it without having to use it. What’s the difference between:

root->right = remover(root->right,valor);

and

root = NULL;?

Aren’t the two trying to modify the pointer without using a pointer to the pointer? So, the two shouldn’t be wrong?

FUNCTION:

Node * remover(Node *root, int valor){
if(root == NULL){
cout<<"Elemento não encontrado."<<endl; 
}
else{
if(root->data == valor){
    root = NULL;
    cout<<"Elemento removido!"<<endl;
    return root;
}
if(root->data > valor){
    /* 1 */ //remover(root->left,valor); não funciona
    /* 2 */root->left = remover(root->left,valor); // funciona
    return root;
}
if(root->data < valor){
    /* 1 */ //remover(root->right,valor);  não funciona
    /* 2 */root->right = remover(root->right,valor); // funciona        
    return root;        
}
}
}

Answer 1

The first line is modifying root->right and not root. The proof of this is that if you display on the screen the address of root before and after this operation, there will be no change.

The second line tries to change (judging by semantics) the value saved in the pointer root. To change this value from the inside of a function, such a function needs the address of the variable that contains this value.

Remember: root simply stores a memory address. Changing what is in the place pointed by that address is not the same as changing that address.