On Sunday 19 August 2001 05:41, you wrote:

Does anyone know how to replace or take out a node from a tree structure ?

Using Haskell, the way we deal with trees is usually rather different from what is done in C or Lisp. Suppose you're dealing with a typical tree in Haskell, defined something like data Tree = Branch Tree Tree | Leaf String An example of a tree structure would be a_tree = Branch (Leaf "x") (Branch (Leaf "x") (Leaf "y")) The usual thing is to define a function which performs the desired replacement, for example replace_first_branch :: Tree -> Tree -> Tree replace_first_branch tree_in new_branch = tree_out where Branch first_branch second_branch = tree_in tree_out = Branch new_branch second_branch or replace_first_leaf :: Tree -> Tree -> Tree replace_first_leaf tree_in new_branch = r tree_in where r (Leaf x) = new_branch r (Branch x y) = Branch (r x) y The above defines a "pure" function, without side effects. The function returns a new tree structure and the input structure is unchanged. But it sounds as if you intend to modify an existing structure. For this purpose you need a structure with modifiable nodes. import IOExts import Monad type TreeRef = IORef Tree data Tree = Branch TreeRef TreeRef | Leaf String replaceFirstLeaf r t = do Branch r1 r2 <- readIORef r is_leaf <- testLeaf r1 when is_leaf (writeIORef r (Branch t r2)) testLeaf r = do v <- readIORef r return $ case v of Leaf x -> True otherwise -> False The above code uses Hugs extensions for modifiable references, i.e. the IORef type, with functions readIORef and writeIORef. Most Haskell implementations support an extension of this kind.