How do I read what's in a binary tree in Haskell?

Question

I have the following type:

data Tree = Branch (Tree) (Tree) | Leaf Int deriving(Show)

now I want to create a function that gives the highest value leaf in the tree. But I am stuck, because I don't know how to get the two following trees of a given tree.

For example I have a tree a that looks like this: let a = Branch (Leaf 10) (Leaf 2)

How do I read the Leaf with value 10 in another function?

Maybe you should try first with the maximum of a plain list instead of a tree. It is the same approach, known as pattern matching together with recursion. Both data structures have exactly two constructors. See function sum' here in LYAH for example. — jpmarinier
– jpmarinier, Commented Jan 23, 2022 at 16:24

Chris · Accepted Answer · 2025-08-27 00:19:45Z

When dealing with recursion, you need to keep in mind a base case.

For:

data Tree = Branch (Tree) (Tree) | Leaf Int deriving(Show)

Your base case (no recursion) is Leaf Int. The maximum value of a leaf is very simply the value held by that leaf.

maxTree :: Tree -> Int
maxTree (Leaf n) = n

Now, how do you deal with a Branch? What is a Branch? It's basically two more Trees.

Consider a very simple Tree:

   Branch
    /  \
   /    \
  /      \
Leaf 1   Leaf 4

The max value is obviously 4. Why? Because the maxTree computation on the right branch was bigger than the maxTree calculation on the left branch.

Consider a more complex Tree:

   Branch
    /  \
   /    \
  /      \
Leaf 1   Branch
          / \
         /   \
        /     \
     Branch   Leaf 8
      / \
     /   \
    /     \
  Leaf 3  Leaf 9

The answer is obviously 9. How do we know this? Well, maxTree tells us the maximum value of Leaf 1 is 1. Each Branch's max value is computed by figuring out the maximum value of the left and right branches. If we call maxTree recursively on each Branch eventually we compare 3 to 9. Clearly 9 is bigger. Now we're comparing 9 to 8. 9 is again bigger. And then 9 to 1, and 9 wins out again.

Now you just need to translate that into code.

maxTree :: Tree -> Int
maxTree (Leaf n) = n
maxTree (Branch left right) = ...

Binary Search Tree

Note that all of this becomes much easier if we're not just talking about a binary tree, but a binary search tree with a strict ordering invariant where the contents of the left branch are always less than the contents of the right branch.

maxBSTree :: BSTree -> Int
maxBSTree (Leaf n) = n
maxBSTree (Branch _ right) = maxBSTree right

Daniel Wagner · Accepted Answer · 2022-01-23 18:43:09Z

6

A typical skeleton looks like this:

maxTree :: Tree -> Int
maxTree (Branch x y) = {- TODO -}
maxTree (Leaf n) = {- TODO -}

answered Jan 23, 2022 at 18:43

Daniel Wagner

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