CODEWITHSUNDEEP
functional-programmingocamlbinary-search-tree
Jackson Tale
Jackson Tale

Reputation: 25812

The correct way to build a Binary Search Tree in OCaml

Ok, I have written a binary search tree in OCaml.

type 'a bstree = 
    |Node of 'a * 'a bstree * 'a bstree
    |Leaf


let rec insert x = function
    |Leaf -> Node (x, Leaf, Leaf)
    |Node (y, left, right) as node -> 
        if x < y then
            Node (y, insert x left, right)
        else if x > y then
            Node (y, left, insert x right)
        else
            node

The above code was said to be good in The right way to use a data structure in OCaml

However, I found a problem. This insert will only work when building a bst from a list in one go, such as 

let rec set_of_list = function
     [] > empty
   | x :: l > insert x (set_of_list l);;

So if we build a bst from a list continuously, no problem, we can get a complete bst which has all nodes from the list.

However, if I have a bst built previously and now I wish to insert a node, then the resulting bst won't have complete nodes from the previous tree, am I right?

Then how should I write a bst in OCaml so that we create a new bst with all nodes from previous tree to keep the previous tree immutable? If every time I need to copy all nodes from old bst, will that impact the performance?

Edit:

So let's say initially, a bst is created with one node t1 = (10, Leaf, Leaf).

then I do let t2 = insert 5 t1, then I get t2 = (10, (5, Leaf, Leaf), Leaf), right? inside t2, let's give a variable c1 to the child node (5, Leaf, Leaf)

then I do let t5 = insert 12 t2, then I get t3 = (10, (5, Leaf, Leaf), (15, Leaf, Leaf)). let's give a variable c2 to the child node (5, Leaf, Leaf)

So my question is whether c1 == c2? Are the two (5, Leaf, Leaf)s in t2 and t3 exactly the same?

Upvotes: 2

Views: 9396

Answers (2)

Jeffrey Scofield
Jeffrey Scofield

Reputation: 66818

I'll try to answer the sharing part of your question. The short answer is yes, the two parts of the two trees will be identical. The reason immutable data works so well is that there are no limitations on the possible sharing. That's why FP works so well.

Here's a session that does what you describe:

# let t1 = Node (10, Leaf, Leaf);;
val t1 : int bstree = Node (10, Leaf, Leaf)
# let t2 = insert 5 t1;;
val t2 : int bstree = Node (10, Node (5, Leaf, Leaf), Leaf)
# let t3 = insert 12 t2;;
val t3 : int bstree = Node (10, Node (5, Leaf, Leaf), Node (12, Leaf, Leaf))
# let Node (_, c1, _) = t2;;
val c1 : int bstree = Node (5, Leaf, Leaf)
# let Node (_, c2, _) = t3;;
val c2 : int bstree = Node (5, Leaf, Leaf)
# c1 == c2;;
- : bool = true

The long answer is that there's no guarantee that the two parts will be identical. If the compiler and/or runtime can see a reason to copy a subtree, it's also free to do that. There are cases (as in distributed processing) where that would be a better choice. Again the great thing about FP is that there are no limitations on sharing, which means that sharing is neither required nor forbidden in such cases.

Upvotes: 5

stonemetal
stonemetal

Reputation: 6208

Look at the accepted answer to the linked question. To be specific this line here:

let tree_of_list l = List.fold_right insert l Leaf

Work out the chain of what is happening. Take the list 1,2,3.

First we have no tree and the result of insert 1 Leaf.

call this T1

Next is the tree generated by insert 2 T1

call this T2

Then the tree generated by insert 3 T2

This is what is returned as the result of Tree_of_list.

If we call the result T3 then somewhere else in code call insert 4 T3 there is no difference in the result returned from insert than in calling Tree_of_list with the list 1,2,3,4.

Upvotes: 2

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