Defining BST::contains()
The stub of BST::contains() should look like this:
bool BST::contains(const Item& item) const {
}
The BST::contains() method should distinguish between the two cases:
 If the BST is empty: return false.
 Otherwise: "pass the buck" by returning whatever
myRoot>contains(item) returns.
Designing Node::contains()
Since a Node is recursively defined,
we might define Node::contains(item) recursively.
One way to design a recursive algorithm for this method is as follows:
Basis. There are three trivial cases:

If item is equal to myItem:

If item belongs in my left subtree and my left subtree is empty:

If item belongs in my right subtree and my right subtree is empty:
Induction Step. There are two cases:

If item belongs in my left subtree and my left subtree is not empty:
 "Pass the buck" to the node in my left subtree and return whatever it returns.

If item belongs in my right subtree and my right subtree is not empty:
 "Pass the buck" to the node in my right subtree and return whatever it returns.
Defining Node::contains()
These observations can be reorganized into the following algorithm for
our Node::contains(item) method:
 If item is less than myItem:
 If myLeft is NULL:
 Otherwise:
 "Pass the buck" by returning whatever
myLeft>contains(item) returns.
 Otherwise, if item is greater than myItem:
 If myRight is NULL:
 Otherwise:
 "Pass the buck" by returning whatever
myRight>contains(item) returns.
 Otherwise (item is equal to myItem):