Binary Search Tree Assignment
We will implement a binary search tree data structure as well as a few Higher Order Function Hall of Fame inductees.
Contents
Background
order
SML's General structure's order
datatype order = LESS | EQUAL | GREATER
Values of type order are used when comparing elements of a type that has a linear ordering.
Functions which take (('a * 'a) -> order) functions behave as Int.compare does:
- compare (i, j)
- returns LESS, EQUAL, or GREATER when i is less than, equal to, or greater than j, respectively.
list
We will implement some of the higher ordered functions list provides on our binary tree.
traversal order
datatype traversal_order = LNR | RNL
Code to Implement
file: | src/main/sml/binary_tree/binary_tree.sml | |
functions: | insert remove find ... |
signature BINARY_SEARCH_TREE = sig type 'k compare_function = (('k * 'k) -> order) type ('a,'k) to_key_function = 'a -> 'k type ('a,'k) tree; val create_empty_tree : ('k compare_function * ('a,'k) to_key_function) -> ('a,'k) tree val create_empty_simple_tree : ('a compare_function) -> ('a,'a) tree val insert : (('a,'k) tree * 'a) -> (('a,'k) tree * 'a option) val remove : (('a,'k) tree * 'k) -> (('a,'k) tree * 'a option) val find : (('a,'k) tree * 'k) -> 'a option val fold_lnr : (('a * 'b) -> 'b) -> ('b) -> (('a,'k) tree) -> 'b val fold_rnl : (('a * 'b) -> 'b) -> ('b) -> (('a,'k) tree) -> 'b val to_string : ('a -> string) -> (('a,'k) tree) -> string end
insert
fun insert (compare_function:(('a * 'a) -> order)) (t:'a tree) (item:'a) : 'a tree = raise NotYetImplemented
remove
Remove contains the most challenging aspect of this studio. When instructed to remove a node from a tree, there are several cases:
not found
What will indicate that you reached the point where you know the node is not found in the tree?
note: this has a trivial solution.
no child in the left tree
How will you detect this pattern?
note: this has a trivial solution.
no child in the right tree
how will you detect this pattern?
note: this has a trivial solution.
no children are present
If you need to remove a node and it has both children, now you have a legit problem. You must maintain a correct binary search tree.
A common approach is to choose one of the following:
- remove the right most descendant in the left child, and promote it to be the node at the current level, or
- remove the left most descendant in right left child, and promote it to be the node at the current level
The image below shows finding the left most child in the right subtree for promotion:
Building a helper function will likely be helpful.
fun remove (equal_less_greater_function:'a -> order) (t:'a tree) : 'a tree = raise NotYetImplemented
find
reference: List.find
fun find (equal_less_greater_function : 'a -> order) (t:'a tree) : 'a option = raise NotYetImplemented
fold_order_hof
reference: List.foldl foldr
note: this function is curried.
fun fold_order_hof (order:traversal_order) (f : 'a * 'b -> 'b) (init : 'b) (t : 'a tree) : 'b = raise NotYetImplemented
fun fold_lnr f init t = fold_order_hof LNR f init t fun fold_rnl f init t = fold_order_hof RNL f init t
Testing
file: | unit_test_binary_search_tree.sml | |
source folder: | src/test/sml/binary_search_tree |
Pledge, Acknowledgments, Citations
file: | studio-binary-search-tree-pledge-acknowledgments-citations.txt |
More info about the Honor Pledge