Am Sonntag 08 November 2009 23:24:59 schrieb spot135:
Hi,
Ok what im trying to do is replace an element in an n-ary and display the completed tree.
So this is what ive got so far.
data Tree a = Empty |Leaf a|Branch a [Tree a] deriving (Show)
replace x y tree should give a tree with the same structure, values equal to x should be replaced by y, all other values left as they are?
replace :: Eq a=>a->a->Tree a -> (Tree a)
replace x y Empty = Element x
That should surely be replace _ _ Empty = Empty ?
replace x y (Leaf z)
| x==z = (Leaf y)
-- | otherwise =
| otherwise = Leaf z
replace x y (Branch z lis)
| x==z = (Branch y lis) |otherwise = replaceA x y clis
You'd want to recur, I believe, use map (replace x y)
replaceA :: Eq a=> a->a->[Tree a]->(Tree a) replaceA _ _ [] = Empty -- run out replaceA x y (h:r)
|Q suc = suc |otherwise = replaceA x y r
where suc=replace x y h
Q :: a -> Bool Q a = True Q a = False
Theres two main problems i am having. One is I'm not sure what should be the "otherwise" in the leaf function. Another problem is it doesn't show the whole tree just the part after the replacement.
If you think this is complete bull I'll quite happily start over if somebody could give me some insight into what is the best way of tackling this problem is.
Many thanks in advance
spot