It’s not clear what the action of `zip` should be on trees with data only at the leaves: if the two trees are mismatched, what should the zipped tree look like? So it is easier to define `zipWith` on an `InternalTree`, and of course to define `zip` in terms of `zipWith`.

```
zipWithInternalTree :: (a -> b -> c) -> InternalTree a
-> InternalTree b -> InternalTree c
zipWithInternalTree f ILeaf _ = ILeaf
zipWithInternalTree f _ ILeaf = ILeaf
zipWithInternalTree f (IBranch a x y) (IBranch b i j)
= IBranch (f a b) x' y'
where x' = (zipWithInternalTree f x i)
y' = (zipWithInternalTree f y j)
zipInternalTree :: InternalTree a -> InternalTree b -> InternalTree (a,b)
zipInternalTree x y = zipWithInternalTree (,) x y
```

## One response to “Exercise 7.4”

Internal and othe trees:

>zipWithIT f (IBranch x l1 r1) (IBranch y l2 r2) =

> IBranch (f x y) (zipWithIT f l1 l2) (zipWithIT f r1 r2)

>zipWithIT _ _ _ = ILeaf

>

>zipWithT f (Leaf a) (Leaf b) = Leaf (f a b)

>zipWithT _ (Branch _ _) (Leaf _) = error “unzipable trees”

>zipWithT _ (Leaf _) (Branch _ _) = error “unzipable trees”

>zipWithT f (Branch l1 r1) (Branch l2 r2) =

> Branch (zipWithT f l1 l2) (zipWithT f r1 r2)

>

>

>zipWithFT f (FBranch x l1 r1) (FBranch y l2 r2) =

> FBranch (f x y) (zipWithFT f l1 l2) (zipWithFT f r1 r2)

>zipWithFT f (FLeaf x ) (FBranch y l r) = FLeaf (f x y)

>zipWithFT f (FBranch y l r) (FLeaf x ) = FLeaf (f x y)

>zipWithFT f (FLeaf y) (FLeaf x ) = FLeaf (f x y)

>

>zipT = zipWithT (,)

>zipIT = zipWithIT (,)

>zipFT = zipWithFT (,)