I'm trying to write code that if given a tree, will go through the tree and return the minimum value in that tree, if the tree is empty, it will return val. What I have right now compiles but will not run. Any help?
minValue :: Ord a => a -> BTree a -> a
minValue val Empty = val
minValue val (BNode v left Empty) = minimum [minValue v left]
minValue val (BNode v Empty right) = minimum [minValue v right]
minValue val (BNode v left right) = minimum ([minValue v left]++[minValue v right])
I'm assuming that BTree is defined as
data BTree a = Empty | BNode a (BTree a) (BTree a) deriving (Eq, Show)
Although for future reference please include data type definitions in your question.
The key to the solution here is that the minimum of a node is the minimum of its value and the mins of each branch:
minValue :: Ord a => a -> BTree a -> a
minValue val Empty = val
minValue val (BNode v left right) =
let leftMin = minValue val left
rightMin = minValue val right
in ???
Instead of worrying if the left or right is Empty, just trust the recursion to handle it. If left is Empty, then minValue val left will just be val, and similarly for right. Then you have 4 values in scope that you want to determine the minimum of, val, v, leftMin, and rightMin. How might you do that?
Related
I have trees of the form:
data Tree a = Leaf | Node (Tree a) a (Tree a)
I created a function to look up the value of a node in a tree based on an in-order, left-to-right traversal.
getElem :: Tree a -> Int -> Maybe a
getElem Leaf _ = Nothing
getElem (Node l x r) n
| s == n = Just x
| n < s = getElem l n
| otherwise = getElem r (n - s - 1)
where
s = size l
I now want to write a method to be able to update a tree. It should be able to take in a tree, an index and a value and update the node at that index with the value. So far I have:
update :: Tree a -> Int -> a -> Tree a
update Leaf _ _ = Leaf
update (Node l x r) index c
| s == index = (Node l c r)
| index < s = update l index c
| otherwise = update r (index - s - 1) c
where
s = size l
This function is able to add but it obviously returns just the added node itself. I want to be able to return the entire tree post the 'update' with the new node or return the tree as is if the index is out of bounds.
Could anybody give me some idea how to proceed with this?
Edit 1:
Okay, I understand that I am basically discarding the remainder of my tree when recursing here. So:
update :: Tree a -> Int -> a -> Tree a
update Leaf _ _ = Leaf
update (Node l x r) index c
| s == index = (Node l c r)
| index < s = update (Node l2 x r) index c
| otherwise = update (Node l x r2) (index - s - 1) c
where
s = size l
l2 = l
r2 = r
Edit 2(Silly me!):
update :: Tree a -> Int -> a -> Tree a
update Leaf _ _ = Leaf
update (Node l x r) index c
| s == index = (Node l c r)
| index < s = (Node (upd l index c) x r)
| otherwise = (Node l x (upd r (index - s - 1) c))
where
s = size l
It took me a bit to wrap my head around it. Thank you for the comments!
Since in Haskell all data is immutable, you can not "update" a tree, you construct a new tree. That tree might however have references to subtrees of the old tree. You thus do not per se construct a completely new tree.
You managed to create an "updated" node, so now the only thing that is missing, is to use the "updated" subtree in a new tree. In that tree, you can use the "old" value together with the other subtree to construct a new one, like:
update :: Tree a -> Int -> a -> Tree a
update Leaf _ _ = Leaf
update (Node l x r) index c
| s == index = Node l c r
| index < s = Node (update l index c) x r
| otherwise = Node l x (update r (index - s - 1) c)
where s = size l
You might also need to change the Leaf case if you "count" a leaf as a node.
Trees with indices are not very common. It might also - in order to boost performance - be better to keep track of the number of items in the left subchild (or both), since then we can just pick the left or right subchild without counting the children. By keeping track of the number, for a complete tree, then updating the tree is an O(log n) operation, not an O(n) operation.
So I have a tree defined as follows:
data Tree = Node Tree Int Tree | Leaf Int
The Int for a Node in this case is the value at that Node. I am trying to check that a tree is balanced, and that the tree is increasing as it's traversed left to right.
To do so I have a recursive function that takes a (Node left x right) and checks that the difference in height of left and right (the nodes below it) is no more than one. I then call balanced again for left and right.
Is it possible to access the Int value of left and right?
Yes, you can write a function that returns the integer at the top node:
getInt (Node _ i _) = i
getInt (Leaf i) = i
E.g.
Prelude> getInt $ Leaf 42
42
Prelude> getInt $ Node (Leaf 42) 123 (Leaf 1337)
123
Of course you can, instead of put variables like left and right, use the constructors again:
Edit, I forget the case of Leaf, it has also an int:
data Tree = Node Tree Int Tree | Leaf Int
exampleSumNodes (Node left x right) = (treeToInt left) + x + (treeToInt right)
treeToInt (Node _ n _) = n
treeToInt (Leaf n ) = n
Assume I have a binary tree:
data Bst a = Empty | Node (Bst a) a (Bst a)
I have to write a function that searches for a value and returns the number of its children. If there is no node with this value, it returns -1. I was trying to write both BFS and DFS, and I failed with both.
Pattern matching is your friend. Your Bst can either be Empty or a Node, so at the toplevel, your search function will be
search Empty = ...
search (Node left x right) = ...
Can an Empty tree possibly contain the target value? With a Node the target value, if present, will be either the node value (x above), in the left subtree, in the right subtree—or perhaps some combination of these.
By “return[ing] the number of its children,” I assume you mean the total number of descendants of the Bst rooted at a Node whose value is the target, which is an interesting combination of problems. You will want another function, say numChildren, whose definition uses pattern matching as above. Considerations:
How many descendants does an Empty tree have?
In the Node case, x doesn’t count because you want descendants. If only you had a function to count the number of children in the left and right subtrees …
Here is a way to do this. Breath-first search can actually be a bit tricky to implement and this solution (findBFS) has aweful complexity (appending to the list is O(n)) but you'll get the gist.
First I have decided to split out the finding functions to return the tree where the node element matches. That simplifies splitting out the counting function. Also, it is easier to return the number of elements than the number of descendants and return -1 in case not found, so the numDesc functions rely on the numElements function.
data Tree a = Empty
| Node a (Tree a) (Tree a)
numElements :: Tree a -> Int
numElements Empty = 0
numElements (Node _ l r) = 1 + numElements l + numElements r
findDFS :: Eq a => a -> Tree a -> Tree a
findDFS _ Empty = Empty
findDFS x node#(Node y l r) | x == y = node
| otherwise = case findDFS x l of
node'#(Node _ _ _) -> node'
Empty -> findDFS x r
findBFS :: Eq a => a -> [Tree a] -> Tree a
findBFS x [] = Empty
findBFS x ((Empty):ts) = findBFS x ts
findBFS x (node#(Node y _ _):ts) | x == y = node
findBFS x ((Node _ l r):ts) = findBFS x (ts ++ [l,r])
numDescDFS :: Eq a => a -> Tree a -> Int
numDescDFS x t = numElements (findDFS x t) - 1
numDescBFS :: Eq a => a -> Tree a -> Int
numDescBFS x t = numElements (findBFS x [t]) - 1
Can someone tell me why this code isn't producing what I want.
data BST = MakeNode BST String BST
| Empty
add :: String -> BST -> BST
add new Empty = (MakeNode Empty new Empty)
add string tree#(MakeNode left value right)
| string > value = MakeNode left value (add string right)
| string < value = MakeNode (add string left) value right
| otherwise = tree
output
"John"
"Doug"
"Charlie"
"Alice"
listToBST :: [String] -> BST
listToBST = foldr add Empty
If we create and function which takes a BST and returns a list in sorted order, modelled after sort . nub, then your Tree is fine as quickcheck tells us. QuickCheck is very easy to use.
import Data.List
import Test.QuickCheck
data BST = MakeNode BST String BST
| Empty
deriving (Show)
add :: String -> BST -> BST
add new Empty = (MakeNode Empty new Empty)
add string tree#(MakeNode left value right)
| string > value = MakeNode left value (add string right)
| string < value = MakeNode (add string left) value right
| otherwise = tree
test = ["alice", "blup", "test", "aa"]
manual_test = inorder (foldr add Empty test) == sort (nub test)
prop_inorder = property inorder_test
where inorder_test :: [String] -> Bool
inorder_test xs = inorder (foldr add Empty xs) == sort (nub xs)
-- return sorted nodes
inorder :: BST -> [String]
inorder (Empty) = []
inorder (MakeNode l x r) = inorder l ++ (x : inorder r)
Just load ghci and then run quickCheck prop_inorder.
Other useful functions are:
reverseOrder :: BST -> [String]
reverseOrder Empty = []
reverseOrder (MakeNode l x r) = reverseOrder r ++ (x : reverseOrder r)
asList :: BST -> [String]
asList Empty = []
asList (MakeNode l x r) = x : (asList l ++ asList r)
And also think about making your tree more general by parameterizing over a:
data BST a = Empty | MakeNode (BST a) a (BST a)
You can make it than an instance of Functor, Monad, Foldable and all kind of handy typeclasses.
I tried it and it seems ok to me. It could help if you gave an example of an input that it doesn't work for.
I think the problem may be that string comparison does not work the way you expect ("123" < "7" because "1" < "7"). If I'm right, you might want to use Ints instead of Strings or even better, the class Ord of all the types that can be ordered using (<).
I'm making a Haskell function to delete a node from a Binary Search Tree.
I know the rules regarding the action needed to be taken depending on the number children
the targeted parent has.
no children - delete,
1 child - replace with the child,
2 children - find the min in the right sub tree and replace the node with the value,
- then, recursively delete the minimum value from the right sub-tree
data BST = MakeNode BST String BST
| Empty
deleteNode :: String -> BST
treeBuilder :: [String] -> BST
treeBuilder = foldr add Empty
add :: String -> BST -> BST
add new Empty = (MakeNode Empty new Empty)
add string tree#(MakeNode left value right)
| string > value = MakeNode left value (add string right)
| string < value = MakeNode (add string left) value right
| otherwise = tree
can't figure out why treeBuilder isn't working correctly either. It just prints Strings Diagonally down to the right.
In these situations, it's best not to think about deleting a node from the tree; it's better to think of how to transform the tree you have into one without the node you want gone.
Let's do some case analysis:
If the tree is empty, then the result is empty, regardless of the key:
delete _ Empty = Empty
If the tree is non-empty, we have to see if the key matches the node. If it does not match, then we need to transform either the left or right subtree based upon whether the key is greater-than or less-than the node:
delete key (MakeNode l key' r) | key < key' = MakeNode (delete key l) key' r
delete key (MakeNode l key' r) | key > key' = MakeNode l key' (delete key r)
If it does match (which it must, since all of the no-match cases have been dealt with), then we have to figure out how to create a new tree without the root node. From your description, if the node has no children, just delete it:
delete _ (MakeNode Empty _ Empty) = Empty
If the node has one child, use that:
delete _ (MakeNode l _ Empty) = l
delete _ (MakeNode Empty _ r) = r
Otherwise, find and delete the minimum key in the right subtree, and use it as the new root's key:
delete _ (MakeNode l _ r) = MakeNode l key r' -- make a new root with min key and new r
where key = minKey r -- find the minimum key in the right subtree
r' = delete key r -- new right subtree with min key removed
-- a helper function to find the minimum key in a tree
-- !! does not work on Empty tree !!
minKey (MakeNode Empty key _) = key
minKey (MakeNode l _ _) = minKey l
You can't! Everything is immutable!
What you can do is make a new tree that's exactly the same as the old one, except with one node removed. (Don't worry, your compiler won't need to duplicate much memory. Remember, everything is immutable. That means that the implementation can safely re-use the common parts!)
As such, your deleteNode function won't be of type String -> BST, it will be of type String -> BST -> BST. The String is the label you want removed, the first BST is the input tree, the second BST is the output tree.
As #Ingo mentioned, you can implement deletion recursively by implementing the function:
deleteNode :: String -> BST -> BST
deleteNode _ Empty = ... -- Handle the empty case
deleteNode x (BST left a right) | x == a = ... -- Delete the node
| x < a = ... -- Recurse on the lesser node
| otherwise = ... -- Recurse on the greater node
If you want to do some general munging beyond deletion (insertion, changes, etc.) in a traversable data structure (trees, lists, etc) I suggest you read up on zippers. They'll help you immensely.
Once you have a zipper for a binary tree, you can use zipper functions to delete nodes in the tree. If you'd like help implementing a zipper for your binary search tree data structure, let me know and I'll expand this. Right now it's probably overkill.
Be warned, a zipper won't re-balance your binary search tree for you. If you want to remove a node from your binary search tree and keep it balanced, that's a whole new can of worms.
There are a number of common balancing algorithms you could use, depending upon your taste. I suggest getting it working in an unbalanced fashion first, and then asking separate questions if you have trouble balancing it.
And, of course, if you want an efficient, out-of-the-box, already-implemented, balancing binary search tree in haskell -- just import Data.Map!
Here is a deletion function implemented in Haskell using Mutual Recursion
The type of the tree is:
type Key = Int
data BST = Nil | Node Key BST BST deriving (Show, Eq)
and here is the delete function:
delete :: Key -> BST -> BST
delete k Nil = Nil
delete k x#(Node a l r)
| (k < a) = Node a (delete k l) r
| (k > a) = Node a l (delete k r)
| (k == a) = delete' k x
delete' :: Key -> BST -> BST
delete' k (Node a l r)
| (l == Nil) = r
| (r == Nil) = l
| otherwise = let (k,t) = maxAndDelete l
in Node k t r
-- This function finds the maximum and then deletes the node as well
maxAndDelete :: BST -> (Key,BST)
maxAndDelete t = let m = treeMaximum t
in (m,delete m t)