[SOLVED] CS代考计算机代写 — Group members:

— Group members:
—* Han-Hsing Pao, 933651943
—* Jiayu Han, 932907189
—* Name, ID
—
— Grading note: 10pts total
—* 2pts each for encodeList and mapTree
—* 3pts each for valueAt and pathTo
module HW2 where

— | Binary trees with nodes labeled by values of an arbitrary type.
data Tree a
= Node a (Tree a) (Tree a)
| End
deriving (Eq,Show)

— | One step in a path, indicating whether to follow the left subtree (L)
— or the right subtree (R).
data Step = L | R
deriving (Eq,Show)

— | A path is a sequence of steps. Each node in a binary tree can be
— identified by a path, indicating how to move down the tree starting
— from the root.
type Path = [Step]

— | Create a leaf node.
leaf :: a -> Tree a
leaf x = Node x End End

— | An example tree.
ex :: Tree Int
ex = Node 4 (Node 3 (leaf 2) End)
(Node 7 (Node 5 End (leaf 6))
(leaf 8))

— | Encode a list as a tree with only right branches.
—
— >>> encodeList []
— End
—
— >>> encodeList [1,2,3,4]
— Node 1 End (Node 2 End (Node 3 End (Node 4 End End)))
—
— >>> encodeList “:-D”
— Node ‘:’ End (Node ‘-‘ End (Node ‘D’ End End))
—
encodeList :: [a]-> Tree a
encodeList [] = End
encodeList (x:xs) = Node x End (encodeList xs)

–If stdin is empty, we just need to output as End, which is the base case.
–Since Encode is a list as a tree with only right branches, the left branch
—is always gonna be End. The format is just (Node x end xs), which is just a simple
—recursion.

— | Map a function over a tree. Applies the given function to every label
— in the tree, preserving the tree’s structure.
—
— >>> mapTree odd End
— End
—
— >>> mapTree even (Node 5 (leaf 2) End)
— Node False (Node True End End) End
—
— >>> (mapTree not . mapTree even) (Node 5 End (leaf 2))
— Node True End (Node False End End)
—
— >>> mapTree (+10) ex
— Node 14 (Node 13 (Node 12 End End) End) (Node 17 (Node 15 End (Node 16 End End)) (Node 18 End End))
—
— >>> ex == (mapTree (subtract 27) . mapTree (+27)) ex
— True
—
mapTree :: (a -> b) -> Tree a -> Tree b
mapTree f End= End
mapTree f (Node x y z) = Node (f x) (mapTree f y) (mapTree f z)

— | Get the value at the node specified by a path. Returns ‘Nothing’ if
— the given path is invalid.
—
— >>> valueAt [] ex
— Just 4
—
— >>> valueAt [L,L] ex
— Just 2
—
— >>> valueAt [L,R] ex
— Nothing
—
— >>> valueAt [R,L,R] ex
— Just 6
—
— >>> valueAt [L,L,L] ex
— Nothing
—
valueAt :: Path -> Tree a -> Maybe a
valueAt _ End = Nothing
valueAt [] (Node x _ _) = Just x
valueAt (L:xs) (Node _ leftside _)= valueAt xs leftside
valueAt (R:xs) (Node _ _ rightside) = valueAt xs rightside

—Since the output is Just[] or Nothing, we need to use Maybe.
—First we need to set the base cases:
—if there is an end, then output will be Nothing
—if the path is empty, then output will be Just x
—else we just need to do recursion based on the indicators from the list
—for either left subtree or right subtree.

— | Find a path to a node that contains the given value.
—
— >>> pathTo 3 (leaf 5)
— Nothing
—
— >>> pathTo 5 ex
— Just [R,L]
—
— >>> pathTo 6 ex
— Just [R,L,R]
—
— >>> pathTo 4 ex
— Just []
—
— >>> pathTo 10 ex
— Nothing
—
pathTo :: Eq a => a -> Tree a -> Maybe Path
pathTo x End= Nothing
pathTo x (Node xs leftside rightside)
| x == xs= Just []
| otherwise= case (pathTo x leftside, pathTo x rightside ) of
(Just p, _) -> Just (L:p)
(_, Just p) -> Just (R:p)
_ -> Nothing

–First, we set the base case for End, which will return Nothing.
–There will be 3 cases:
—x == xs, returns Just []
— search left subtree
—search right subtree
–We use pair to group both left and right recursion together, which add L/R to
—the current path list till the value is found, otherwise the value is not found
—after recursion, return Nothing.