void instert(NODE**root, int value)
{
...
insert(&(*root)->left,value);
...
}
void search(NODE*root, int value)
{
...
search(root->left, value);
...
}
Why we use "&" here: insert(&(*root)->left,value);
But we do not use "&" here: search(root->left, value);
An extra level of indirection is added to insert
function so that it could modify the pointer. This is not necessary in case of the search
function, because it never modifies the pointer passed to it.
Specifically, there needs to be a place in the insert
function that does something like this:
*root = malloc(sizeof(NODE));
(*root)->left = NULL;
(*root)->right = NULL;
(*root)->value = value;
This would modify the pointer which is pointed to by the pointer to pointer.
Note that it is possible to avoid this extra level of indirection by returning the new value of the pointer from insert
, like this:
NODE* insert(NODE*root, int value) {
...
root->left = insert(root->left, value);
...
}
However, this changes the way in which all callers must call insert
, including the top-level caller: rather than writing
insert(&root, value);
he would be forced to write
root = insert(root, value);
The expression:
*root->left
Is equivalent to:
*(root->left)
Due to operator precedence.
So you need:
(*root)->left
If you want the left
member that *root
points to.
And then:
&(*root)->left
Is the pointer to the left
member of *root
, which is then of type NODE **
, what the insert
function requires.
&(*
isn't a statement.
It is parts of the &
-operator being applied to the expression (*root)->left
.