Writing a function to compute the left spine of a tree - haskell

I'm needing some help in figuring out how to create a leftSpine function in Haskell.
Basically, it is supposed to take all the left most leafs and put them into a list, but whenever I run my code I get an empty list. Any help would be appreciated.
Here is my code.
data Tree x = Leaf | Node (Tree x) x (Tree x)
deriving Show
leftSpine :: Tree x -> [x]
leftSpine Leaf = []
leftSpine (Node lt x rt) = (leftSpine lt)
And here is my code to test it.
leftSpine (Node (Node (Node Leaf 1 Leaf) 2 (Node Leaf 3 Leaf))
4
(Node (Node Leaf 5 Leaf) 6 (Node Leaf 7 Leaf)))
It should equal [4,2,1] but it just comes up as [].

leftSpine :: Tree x -> [x]
leftSpine Leaf = []
leftSpine (Node lt x rt) = x:leftSpine lt
You weren't actually putting anything into the list. The difference is the x:leftSpine lt instead of leftSpine lt.

Your code plainly says that the result is always the empty list.
The first case says that the left spine of a leaf is the empty list. Ok so far.
The second case says that the left spine of a node is exactly the left spine of the node's left child.
So if we want to find the left spine of a tree, we'll just keep chasing down the left children of the nodes we reach, knowing that the answer is exactly equal to the left spine of the next left child. Either we eventually find a leaf, and our result is the empty list, or the tree is infinite and we never return a result at all. There is nothing in your code that could ever return anything else.
The key to writing a recursive function of this kind is to figure out what the answer is for the base case (Leaf, here), and then for the non-base cases (Node, here) you need to figure out how to combine the sub-solutions with the local information here to generate a full solution.
In this case the left spine of a Leaf is easy, as there's no data at all. So then how would you combine the information from Node lt x rt with leftSpine lt to get the left spine of the whole tree?

Related

Building a Binary Tree (not BST) in Haskell Breadth-First

I recently started using Haskell and it will probably be for a short while. Just being asked to use it to better understand functional programming for a class I am taking at Uni.
Now I have a slight problem I am currently facing with what I am trying to do. I want to build it breadth-first but I think I got my conditions messed up or my conditions are also just wrong.
So essentially if I give it
[“A1-Gate”, “North-Region”, “South-Region”, “Convention Center”, “Rectorate”, “Academic Building1”, “Academic Building2”] and [0.0, 0.5, 0.7, 0.3, 0.6, 1.2, 1.4, 1.2], my tree should come out like
But my test run results are haha not what I expected. So an extra sharp expert in Haskell could possibly help me spot what I am doing wrong.
Output:
*Main> l1 = ["A1-Gate", "North-Region", "South-Region", "Convention Center",
"Rectorate", "Academic Building1", "Academic Building2"]
*Main> l3 = [0.0, 0.5, 0.7, 0.3, 0.6, 1.2, 1.4, 1.2]
*Main> parkingtree = createBinaryParkingTree l1 l3
*Main> parkingtree
Node "North-Region" 0.5
(Node "A1-Gate" 0.0 EmptyTree EmptyTree)
(Node "Convention Center" 0.3
(Node "South-Region" 0.7 EmptyTree EmptyTree)
(Node "Academic Building2" 1.4
(Node "Academic Building1" 1.2 EmptyTree EmptyTree)
(Node "Rectorate" 0.6 EmptyTree EmptyTree)))
A-1 Gate should be the root but it ends up being a child with no children so pretty messed up conditions.
If I could get some guidance it would help. Below is what I've written so far::
data Tree = EmptyTree | Node [Char] Float Tree Tree deriving (Show,Eq,Ord)
insertElement location cost EmptyTree =
Node location cost EmptyTree EmptyTree
insertElement newlocation newcost (Node location cost left right) =
if (left == EmptyTree && right == EmptyTree)
then Node location cost (insertElement newlocation newcost EmptyTree)
right
else if (left == EmptyTree && right /= EmptyTree)
then Node location cost (insertElement newlocation newcost EmptyTree)
right
else if (left /= EmptyTree && right == EmptyTree)
then Node location cost left
(insertElement newlocation newcost EmptyTree)
else Node newlocation newcost EmptyTree
(Node location cost left right)
buildBPT [] = EmptyTree
--buildBPT (xs:[]) = insertElement (fst xs) (snd xs) (buildBPT [])
buildBPT (x:xs) = insertElement (fst x) (snd x) (buildBPT xs)
createBinaryParkingTree a b = buildBPT (zip a b)
Thank you for any guidance that might be provided. Yes I have looked at some of the similar questions I do think my problem is different but if you think a certain post has a clear answer that will help I am willing to go and take a look at it.
Here's a corecursive solution.
{-# bft(Xs,T) :- bft( Xs, [T|Q], Q). % if you don't read Prolog, see (*)
bft( [], Nodes , []) :- maplist( =(empty), Nodes).
bft( [X|Xs], [N|Nodes], [L,R|Q]) :- N = node(X,L,R),
bft( Xs, Nodes, Q).
#-}
data Tree a = Empty | Node a (Tree a) (Tree a) deriving Show
bft :: [a] -> Tree a
bft xs = head nodes -- Breadth First Tree
where
nodes = zipWith g (map Just xs ++ repeat Nothing) -- values and
-- Empty leaves...
(pairs $ tail nodes) -- branches...
g (Just x) (lt,rt) = Node x lt rt
g Nothing _ = Empty
pairs ~(a: ~(b:c)) = (a,b) : pairs c
{-
nodes!!0 = g (Just (xs!!0)) (nodes!!1, nodes!!2) .
nodes!!1 = g (Just (xs!!1)) (nodes!!3, nodes!!4) . .
nodes!!2 = g (Just (xs!!2)) (nodes!!5, nodes!!6) . . . .
................ .................
-}
nodes is the breadth-first enumeration of all the subtrees of the result tree. The tree itself is the top subtree, i.e., the first in this list. We create Nodes from each x in the input xs, and when the input
is exhausted we create Emptys by using an indefinite number of Nothings instead (the Empty leaves' true length is length xs + 1 but we don't need to care about that).
And we didn't have to count at all.
Testing:
> bft [1..4]
Node 1 (Node 2 (Node 4 Empty Empty) Empty) (Node 3 Empty Empty)
> bft [1..10]
Node 1
(Node 2
(Node 4
(Node 8 Empty Empty)
(Node 9 Empty Empty))
(Node 5
(Node 10 Empty Empty)
Empty))
(Node 3
(Node 6 Empty Empty)
(Node 7 Empty Empty))
How does it work: the key is g's laziness, that it doesn't force lt's nor rt's value, while the tuple structure is readily served by -- very lazy in its own right -- pairs. So both are just like the not-yet-set variables in that Prolog pseudocode(*), when served as 2nd and 3rd arguments to g. But then, for the next x in xs, the node referred to by this lt becomes the next invocation of g's result.
And then it's rt's turn, etc. And when xs end, and we hit the Nothings, g stops pulling the values from pairs's output altogether. So pairs stops advancing on the nodes too, which is thus never finished though it's defined as an unending stream of Emptys past that point, just to be on the safe side.
(*) Prolog's variables are explicitly set-once: they are allowed to be in a not-yet-assigned state. Haskell's (x:xs) is Prolog's [X | Xs].
The pseudocode: maintain a queue; enqueue "unassigned pointer"; for each x in xs: { set pointer in current head of the queue to Node(x, lt, rt) where lt, rt are unassigned pointers; enqueue lt; enqueue rt; pop queue }; set all pointers remaining in queue to Empty; find resulting tree in the original head of the queue, i.e. the original first "unassigned pointer" (or "empty box" instead of "unassigned pointer" is another option).
This Prolog's "queue" is of course fully persistent: "popping" does not mutate any data structure and doesn't change any outstanding references to the queue's former head -- it just advances the current pointer into the queue. So what's left in the wake of all this queuing, is the bfs-enumeration of the built tree's nodes, with the tree itself its head element -- the tree is its top node, with the two children fully instantiated to the bottom leaves by the time the enumeration is done.
Update: #dfeuer came up with much simplified version of it which is much closer to the Prolog original (that one in the comment at the top of the post), that can be much clearer. Look for more efficient code and discussion and stuff in his post. Using the simple [] instead of dfeuer's use of the more efficient infinite stream type data IS a = a :+ IS a for the sub-trees queue, it becomes
bftree :: [a] -> Tree a
bftree xs = t
where
t : q = go xs q
go [] _ = repeat Empty
go (x:ys) ~(l : ~(r : q)) = Node x l r : go ys q
---READ-- ----READ---- ---WRITE---
{-
xs = [ x x2 x3 x4 x5 x6 x7 x8 … ]
(t:q) = [ t l r ll lr rl rr llr … Empty Empty … … ]
-}
For comparison, the opposite operation of breadth-first enumeration of a tree is
bflist :: Tree a -> [a]
bflist t = [x | Node x _ _ <- q]
where
q = t : go 1 q
go 0 _ = []
go i (Empty : q) = go (i-1) q
go i (Node _ l r : q) = l : r : go (i+1) q
-----READ------ --WRITE--
How does bftree work: t : q is the list of the tree's sub-trees in breadth-first order. A particular invocation of go (x:ys) uses l and r before they are defined by subsequent invocations of go, either with another x further down the ys, or by go [] which always returns Empty. The result t is the very first in this list, the topmost node of the tree, i.e. the tree itself.
This list of tree nodes is created by the recursive invocations of go at the same speed with which the input list of values xs is consumed, but is consumed as the input to go at twice that speed, because each node has two child nodes.
These extra nodes thus must also be defined, as Empty leaves. We don't care how many are needed and simply create an infinite list of them to fulfill any need, although the actual number of empty leaves will be one more than there were xs.
This is actually the same scheme as used in computer science for decades for array-backed trees where tree nodes are placed in breadth-first order in a linear array. Curiously, in such setting both conversions are a no-op -- only our interpretation of the same data is what's changing, our handling of it, how are we interacting with / using it.
Update: the below solution is big-O optimal and (I think) pretty easy to understand, so I'm leaving it here in case anyone's interested. However, Will Ness's solution is much more beautiful and, especially when optimized a bit, can be expected to perform better in practice. It is much more worthy of study!
I'm going to ignore the fake edge labels for now and just focus on the core of what's happening.
A common pattern in algorithm design is that it's sometimes easier to solve a more general problem. So instead of trying to build a tree, I'm going to look at how to build a forest (a list of trees) with a given number of trees. I'll make the node labels polymorphic to avoid having to think about what they look like; you can of course use the same building technique with your original tree type.
data Tree a = Empty | Node a (Tree a) (Tree a)
-- Built a tree from a breadth-first list
bft :: [a] -> Tree a
bft xs = case dff 1 xs of
[] -> Empty
[t] -> t
_ -> error "something went wrong"
-- Build a forest of nonempty trees.
-- The given number indicates the (maximum)
-- number of trees to build.
bff :: Int -> [a] -> [Tree a]
bff _ [] = []
bff n xs = case splitAt n xs of
(front, rear) -> combine front (bff (2 * n) rear)
where
combine :: [a] -> [Tree a] -> [Tree a]
-- you write this
Here's a full, industrial-strength, maximally lazy implementation. This is the most efficient version I've been able to come up with that's as lazy as possible. A slight variant is less lazy but still works for fully-defined infinite inputs; I haven't tried to test which would be faster in practice.
bft' :: [a] -> Tree a
bft' xs = case bff 1 xs of
[] -> Empty
[t] -> t
_ -> error "whoops"
bff' :: Int -> [a] -> [Tree a]
bff' !_ [] = []
bff' n xs = combine n xs (bff (2 * n) (drop n xs))
where
-- The "take" portion of the splitAt in the original
-- bff is integrated into this version of combine. That
-- lets us avoid allocating an intermediate list we don't
-- really need.
combine :: Int -> [a] -> [Tree a] -> [Tree a]
combine 0 !_ ~[] = [] -- These two lazy patterns are just documentation
combine _k [] ~[] = []
combine k (y : ys) ts = Node y l r : combine (k - 1) ys dropped
where
(l, ~(r, dropped)) = case ts of -- This lazy pattern matters.
[] -> (Empty, (Empty, []))
t1 : ts' -> (t1, case ts' of
[] -> (Empty, [])
t2 : ts'' -> (t2, ts''))
For the less-lazy variant, replace (!l, ~(!r, dropped)) with (!l, !r, dropped) and adjust the RHS accordingly.
For true industrial strength, forests should be represented using lists strict in their elements:
data SL a = Cons !a (SL a) | Nil
And the pairs in the above (l, ~(r, dropped)) should both be represented using a type like
data LSP a b = LSP !a b
This should avoid some (pretty cheap) run-time checks. More importantly, it makes it easier to see where things are and aren't getting forced.
The method that you appear to have chosen is to build the tree up backwards: from bottom-to-top, right-to-left; starting from the last element of your list. This makes your buildBPT function look nice, but requires your insertElement to be overly complex. To construct a binary tree in a breadth-first fashion this way would require some difficult pivots at every step past the first three.
Adding 8 nodes to the tree would require the following steps (see how the nodes are inserted from last to first):
. 4
6 6
8 7 8 . .
. .
3
7 4 5
8 . 6 7 8 .
6 2
7 8 3 4
5 6 7 8
5
6 7 1
8 . . . 2 3
4 5 6 7
8 . . . . . . .
If, instead, you insert the nodes left-to-right, top-to-bottom, you end up with a much simpler solution, requiring no pivoting, but instead some tree structure introspection. See the insertion order; at all times, the existing values remain where they were:
. 1
2 3
1 4 5 . .
. .
1
1 2 3
2 . 4 5 6 .
1 1
2 3 2 3
4 5 6 7
1
2 3 1
4 . . . 2 3
4 5 6 7
8 . . . . . . .
The insertion step has an asymptotic time complexity on the order of O(n^2) where n is the number of nodes to insert, as you are inserting the nodes one-by-one, and then iterating the nodes already present in the tree.
As we insert left-to-right, the trick is to check whether the left sub-tree is complete:
if it is, and the right sub-tree is not complete, then recurse to the right.
if it is, and the right sub-tree is also complete, then recurse to the left (starting a new row).
if it is not, then recurse to the left.
Here is my (more generic) solution:
data Tree a = Leaf | Node a (Tree a) (Tree a)
deriving (Eq, Show)
main = do
let l1 = ["A1-Gate", "North-Region", "South-Region", "Convention Center",
"Rectorate", "Academic Building1", "Academic Building2"]
let l2 = [0.0, 0.5, 0.7, 0.3, 0.6, 1.2, 1.4, 1.2]
print $ treeFromList $ zip l1 l2
mkNode :: a -> Tree a
mkNode x = Node x Leaf Leaf
insertValue :: Tree a -> a -> Tree a
insertValue Leaf y = mkNode y
insertValue (Node x left right) y
| isComplete left && nodeCount left /= nodeCount right = Node x left (insertValue right y)
| otherwise = Node x (insertValue left y) right
where nodeCount Leaf = 0
nodeCount (Node _ left right) = 1 + nodeCount left + nodeCount right
depth Leaf = 0
depth (Node _ left right) = 1 + max (depth left) (depth right)
isComplete n = nodeCount n == 2 ^ (depth n) - 1
treeFromList :: (Show a) => [a] -> Tree a
treeFromList = foldl insertValue Leaf
EDIT: more detailed explanation:
The idea is to remember in what order you insert nodes: left-to-right first, then top-to-bottom. I compressed the different cases in the actual function, but you can expand them into three:
Is the left side complete? If not, then insert to the left side.
Is the right side as complete as the left side, which is complete? If not, then insert to the right side.
Both sides are full, so we start a new level by inserting to the left side.
Because the function fills the nodes up from left-to-right and top-to-bottom, then we always know (it's an invariant) that the left side must fill up before the right side, and that the left side can never be more than one level deeper than the right side (nor can it be shallower than the right side).
By following the growth of the second set of example trees, you can see how the values are inserted following this invariant. This is enough to describe the process recursively, so it extrapolates to a list of any size (the recursion is the magic).
Now, how do we determine whether a tree is 'complete'? Well, it is complete if it is perfectly balanced, or if – visually – its values form a triangle. As we are working with binary trees, then the base of the triangle (when filled) must have a number of values equal to a power of two. More specifically, it must have 2^(depth-1) values. Count for yourself in the examples:
depth = 1 -> base = 1: 2^(1-1) = 1
depth = 2 -> base = 2: 2^(2-1) = 2
depth = 3 -> base = 4: 2^(3-1) = 4
depth = 4 -> base = 8: 2^(4-1) = 8
The total number of nodes above the base is one less than the width of the base: 2^(n-1) - 1. The total number of nodes in the complete tree is therefore the number of nodes above the base, plus those of the base, so:
num nodes in complete tree = 2^(depth-1) - 1 + 2^(depth-1)
= 2 × 2^(depth-1) - 1
= 2^depth - 1
So now we can say that a tree is complete if it has exactly 2^depth - 1 non-empty nodes in it.
Because we go left-to-right, top-to-bottom, when the left side is complete, we move to the right, and when the right side is just as complete as the left side (meaning that it has the same number of nodes, which is means that it is also complete because of the invariant), then we know that the whole tree is complete, and therefore a new row must be added.
I originally had three special cases in there: when both nodes are empty, when the left node is empty (and therefore so was the right) and when the right node is empty (and therefore the left could not be). These three special cases are superseded by the final case with the guards:
If both sides are empty, then countNodes left == countNodes right, so therefore we add another row (to the left).
If the left side is empty, then both sides are empty (see previous point).
If the right side is empty, then the left side must have depth 1 and node count 1, meaning that it is complete, and 1 /= 0, so we add to the right side.

Find all the partial paths in a tree

I am trying to create a Haskell function that generates a list containing all the partial paths of a tree, given the tree implementation data Tree a = Empty | Node a (Tree a) (Tree a). For example, if I have a tree
tree = Node 5 (Node 3 Empty Empty ) (Node 2 Empty Empty )
I want to get
[[],[5],[5,3],[5,2]]
How could I make such a function?
First let us consider the type of this function, it must be Tree a -> [[a]].
So what can we do given a node Node x left right? We have the path that is just the node itself - this is just [x], as well as the paths that go through this node to the left- and right sub tree. The paths going throu the left and right sub tree are just what we get if we apply our function to left and right respectively. We now just need to add x to the start of each of those paths and we do that by calling map(x:) paths. (And for an empty one we should get an empty list as there is no path.)
data Tree a = Empty | Node a (Tree a) (Tree a)
tree = Node 5 (Node 3 Empty Empty ) (Node 2 Empty Empty )
pp :: Tree a -> [[a]]
pp (Node x left right) = [[x]] ++ map(x:)(pp left) ++ map(x:)(pp right)
pp Empty = []
Now this has one flaw that the empty path is not considered as a partial path by this function. But we can easily amend that by adding it and wrapping it in another function:
partialPaths :: Tree a -> [[a]]
partialPaths t = [[]] ++ pp t
main = print $ partialPaths tree
Try it online!

Tree to a list of possible paths (values)

Given a tree data structure defined as
data Tree = Node Int Tree Tree | Leaf
How can one transform it into a list of values along all paths?
For example Node 1 (Node 2 Leaf Leaf) (Node 3 Leaf Leaf) should translate to [[1,2], [1,3]].
You have a recursive data structure, so you should expect a recursive solution. The first step in such a solution is to identify the base cases. For this problem, there is the obvious base case: Leaf. We probably also want to treat Node x Leaf Leaf as a base case to avoid duplicate paths.
Now let's write down the type signature. This should make it clear what our base cases should produce.
paths :: Tree -> [[Int]]
For Leaf the sensible thing to do is return a list containing an empty list since a Leaf is representing an empty path.
paths Leaf = [[]]
For Node x Leaf Leaf, we can return a list consisting of a list containing x.
paths (Node x Leaf Leaf) = [[x]]
The next part requires the most thought. We need to consider what to do with the non-base case Node x left right. The strategy here is to assume that we have the result of paths left and paths right and then decide what we need to do with x. We're building paths, so we need to tack x onto the front of all the left paths and the right paths. We also need to combine the two lists into a single list.
paths (Node x left right) = map (x:) (paths left ++ paths right)
And that's it. Of course, now you might want to see if there's a more efficient way to implement it, or if there's a common pattern here instead (i.e. can we write this as a fold?).
The answer here is very simple, so for a Tree structure such as the one you provided, if we have:
data Tree = Node Int Tree Tree | Leaf
We would need a nested function within our function, basically a function which returns a list from a branch, this way we can have the first argument of the Tree (the trunk or initial value) separated from values of either of the branches, but I'm sure there is a more efficient implementation, so here it is:
traverse :: Tree a -> [[Int]] -- Do not use "a" in the return type
traverse Leaf = [[]]
traverse (Node value left right) = [value] ++ treeToList left : [value] ++ treeToList right : []
where treeToList Leaf = []
treeToList (Node a left right) = [a] ++ treeToList left ++ treeToList right
returns
> traverse (Node 2 (Node 3 Leaf Leaf) (Node 4 Leaf Leaf))
> [[2,3],[2,4]]
Be careful with the return type as Tree a denotes that a is polymorphic but the constructor Node takes only values of type Int, denoting the returned list can only be of type Int.

TreeToList in Haskell function

I have this function that i wrote to pass from a tree to a list in Haskell:
treeToList :: Tree -> [Int]
treeToList (Leaf x) = [x]
treeToList (Node left x right) = treeToList left ++ [x] ++ treeToList right
This works just fine, however i have a doubt:
With the input:
treeToList (Node (Leaf 1) 2 (Node (Leaf 3) 4 (Leaf 5)))
the function produces this list:
[1,2,3,4,5]
which, i think is wrong, because if i want to go the other way around, and write the tree from the list, i'm going to write it wrong.
How can i fix this ?
This is a correct way to transform a binary tree of Int into a list by "in-order" traversal. This is the most common way to turn a search tree into a list. There are, however, other valid ways to traverse trees which will produce lists in different orders for various purposes, as you can see on Wikipedia. As András Kovács and chi have indicated, there is generally no way to go backwards, unless you have specific information about the shape of the tree you wish to construct.

Assign Consecutive Numbers to Elements in a BST

So I am trying to add consecutive numbers to the elements in a BST strictly using recursion (no standard prelude functions). Here is what I have so far:
data Tree a = Empty | Node a (Tree a) (Tree a) deriving (Show)
leaf x = Node x Empty Empty
number' :: Int -> Tree a -> Tree (Int, a)
number' a Empty = Empty
number' a (Node x xl xr) = Node (a,x) (number' (a+1) xl) (number' (a+1) xr)
number :: Tree a -> Tree (Int, a)
number = number' 1
number' is an auxiliary function that carries around "a" as a counter. It should add 1 to each recursive call, so I am not sure why it is doing what it is doing.
As of now the level of the element is assigned to each element. I would like the first element to be assigned 1, the element to the left of that 2, the element to the left of that 3, etc. Each element should get a+1 assigned to it and no number should be repeated. Thanks in advance.
I want to first explain why the code in the question assigns level numbers. This will lead us directly to two different solutions, one passed on caching, one based on doing two traversals at once. Finally, I show how the second solution relates to the solutions provided by other answers.
What has to be changed in the code from the question?
The code in the question assigns the level number to each node. We can understand why the code behaves like that by looking at the recursive case of the number' function:
number' a (Node x xl xr) = Node (a,x) (number' (a+1) xl) (number' (a+1) xr)
Note that we use the same number, a + 1, for both recursive calls. So the root nodes in both subtrees will get assigned the same number. If we want each node to have a different number, we better pass different numbers to the recursive calls.
What number should we pass to the recursive call?
If we want to assign the numbers according to a left-to-right pre-order traversal, then a + 1 is correct for the recursive call on the left subtree, but not for the recursive call on the right subtree. Instead, we want to leave out enough numbers to annotate the whole left subtree, and then start annotating the right subtree with the next number.
How many numbers do we need to reserve for the left subtree? That depends on the subtree's size, as computed by this function:
size :: Tree a -> Int
size Empty = 0
size (Node _ xl xr) = 1 + size xl + size xr
Back to the recursive case of the number' function. The smallest number annotated somewhere in the left subtree is a + 1. The biggest number annotated somewhere in the left subtree is a + size xl. So the smallest number available for the right subtree is a + size xl + 1. This reasoning leads to the following implementation of the recursive case for number' that works correctly:
number' :: Int -> Tree a -> Tree (Int, a)
number' a Empty = Empty
number' a (Node x xl xr) = Node (a,x) (number' (a+1) xl) (number' (a + size xl + 1) xr)
Unfortunately, there is a problem with this solution: It is unnecessarily slow.
Why is the solution with size slow?
The function size traverses the whole tree. The function number' also traverses the whole tree, and it calls size on all left subtrees. Each of these calls will traverse the whole subtree. So overall, the function size gets executed more than once on the same node, even though it always returns the same value, of course.
How can we avoid traversing the tree when calling size?
I know two solutions: Either we avoid traversing the tree in the implementation of size by caching the sizes of all trees, or we avoid calling size in the first place by numbering the nodes and computing the size in one traversal.
How can we compute the size without traversing the tree?
We cache the size in every tree node:
data Tree a = Empty | Node Int a (Tree a) (Tree a) deriving (Show)
size :: Tree a -> Int
size Empty = 0
size (Node n _ _ _) = n
Note that in the Node case of size, we just return the cached size. So this case is not recursive, and size does not traverse the tree, and the problem with our implementation of number' above goes away.
But the information about the size has to come from somewhere! Everytime we create a Node, we have to provide the correct size to fill the cache. We can lift this task off to smart constructors:
empty :: Tree a
empty = Empty
node :: a -> Tree a -> Tree a -> Tree a
node x xl xr = Node (size xl + size xr + 1) x xl xr
leaf :: a -> Tree a
leaf x = Node 1 x Empty Empty
Only node is really necessary, but I added the other two for completeness. If we always use one of these three functions to create a tree, the cached size information will always be correct.
Here is the version of number' that works with these definitions:
number' :: Int -> Tree a -> Tree (Int, a)
number' a Empty = Empty
number' a (Node _ x xl xr) = node (a,x) (number' (a+1) xl) (number' (a + size xl + 1) xr)
We have to adjust two things: When pattern matching on Node, we ignore the size information. And when creating a Node, we use the smart constructor node.
That works fine, but it has the drawback of having to change the definition of trees. On the one hand, caching the size might be a good idea anyway, but on the other hand, it uses some memory and it forces the trees to be finite. What if we want to implement a fast number' without changing the definition of trees? This brings us to the second solution I promised.
How can we number the tree without computing the size?
We cannot. But we can number the tree and compute the size in a single traversal, avoiding the multiple size calls.
number' :: Int -> Tree a -> (Int, Tree (Int, a))
Already in the type signature, we see that this version of number' computes two pieces of information: The first component of the result tuple is the size of the tree, and the second component is the annotated tree.
number' a Empty = (0, Empty)
number' a (Node x xl xr) = (sl + sr + 1, Node (a, x) yl yr) where
(sl, yl) = number' (a + 1) xl
(sr, yr) = number' (a + sl + 1) xr
The implementation decomposes the tuples from the recursive calls and composes the components of the result. Note that sl is like size xl from the previous solution, and sr is like size xr. We also have to name the annotated subtrees: yl is the left subtree with node numbers, so it is like number' ... xl in the previous solution, and yr is the right subtree with node numbers, so it is like number' ... xr in the previous solution.
We also have to change number to only return the second component of the result of number':
number :: Tree a -> Tree (Int, a)
number = snd . number' 1
I think that in a way, this is the clearest solution.
What else could we improve?
The previous solution works by returning the size of the subtree. That information is then used to compute the next available node number. Instead, we could also return the next available node number directly.
number' a Empty = (a, Empty)
number' a (Node x xl xr) = (ar, Node (a, x) yl yr) where
(al, yl) = number' (a + 1) xl
(ar, yr) = number' al xr
Note that al is like a + sl + 1 in the previous solution, and ar is like a + sl + sr + 1. Clearly, this change avoids some additions.
This is essentially the solution from Sergey's answer, and I would expect that this is the version most Haskellers would write. You could also hide the manipulations of a, al and ar in a state monad, but I don't think that really helps for such a small example. The answer by Ankur shows how it would look like.
data Tree a = Empty | Node a (Tree a) (Tree a) deriving (Show)
number :: Tree a -> Tree (Int, a)
number = fst . number' 1
number' :: Int -> Tree a -> (Tree (Int, a), Int)
number' a Empty = (Empty, a)
number' a (Node x l r) = let (l', a') = number' (a + 1) l
(r', a'') = number' a' r
in (Node (a, x) l' r', a'')
*Tr> let t = (Node 10 (Node 20 (Node 30 Empty Empty) (Node 40 Empty Empty)) (Node 50 (Node 60 Empty Empty) Empty))
*Tr> t
Node 10 (Node 20 (Node 30 Empty Empty) (Node 40 Empty Empty)) (Node 50 (Node 60 Empty Empty) Empty)
*Tr> number t
Node (1,10) (Node (2,20) (Node (3,30) Empty Empty) (Node (4,40) Empty Empty)) (Node (5,50) (Node (6,60) Empty Empty) Empty)
As suggested by comments in your question that each call to number should return a integer also which needs to be further used for next set of nodes. This makes the signature of the function to:
Tree a -> Int -> (Tree (Int,a), Int)
Looking at the last part of it, it looks like a candidate for State monad i.e state -> (Val,state).
Below code shows how you can do this using State monad.
import Control.Monad.State
data Tree a = Empty | Node a (Tree a) (Tree a) deriving (Show)
myTree :: Tree String
myTree = Node "A" (Node "B" (Node "D" Empty Empty) (Node "E" Empty Empty)) (Node "C" (Node "F" Empty Empty) (Node "G" Empty Empty))
inc :: State Int ()
inc = do
i <- get
put $ i + 1
return ()
number :: Tree a -> State Int (Tree (Int,a))
number Empty = return Empty
number (Node x l r) = do
i <- get
inc
l' <- number l
r' <- number r
return $ Node (i,x) l' r'
main = do
putStrLn $ show (fst (runState (number myTree) 1))

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