Question

How do you delete the minimum in BST? I can’t seem to find a way to keep the tree

data BST = EmptyT
         | Node Float BST BST
           deriving (Eq, Show, Read)

deleteMin :: BST -> Maybe BST
deleteMin EmptyT  = Nothing
deleteMin (Node x left right)
    |left == EmptyT = ? 
    |otherwise = ?

do i need getters and setters?

Answer 1

Haskell does not have “getters and setter” in the OOP sense, though you can
come up with similar concepts. If you wish to delete a value in your binary
tree, you have to construct a new tree with the value missing. That is how you
“keep the tree.”

Assuming you are using a standard BST, then the leftmost node in the tree will
contain the minimum element. So, by traversing your tree towards the left,
you should eventually run into a situation that looks like Node x EmptyT r.
Any other node, you just recursively call deleteMin on the left branch.

This gives a function that looks like

deleteMin :: BST -> Maybe BST
deleteMin EmptyT                = Nothing
deleteMin (Node x EmptyT right) = Just right
deleteMin (Node x left right)   =
    case deleteMin left of
         Nothing -> Nothing
         Just nl -> Just $ Node x nl right

You have to check the result of each call to deleteMin to check for
Nothing. I don’t think you really need to return a Maybe BST, unless you
really need to indicate that there is no element to delete. It makes more
sense (to me at least) to just return an empty tree if there is nothing to
delete.

I think most would also consider the use of pattern matching preferable over using guards with an equality check.