This post discusses a O(n) algorithm that construct a balanced binary search tree (BST) from a sorted list. For instance, consider that we are given a sorted list: [1,2,3,4,5,6,7]
. We have to construct a balanced BST as follows.
4 | 2 6 | | 1 3 5 7
To do so, we use the following definition of Tree
, described in Scala By Example book.
abstract class IntSet
case object Empty extends IntSet
case class NonEmpty(elem: Int, left: IntSet, right: IntSet) extends IntSet
One straight-forward approach would be to repeatedly perform binary search on the given list to find the median of the list, and then, to construct a balanced BST recursively. Complexity of such approach is O(nlogn), where n is the number of elements in the given list.
A better algorithm constructs balanced BST while iterating the list only once. It begins with the leaf nodes and construct the tree in a bottom-up manner. As such, it avoids repeated binary searches and achieves better runtime complexity (i.e., O(n), where n is the total number of elements in the given list). Following Scala code outlines this algorithm, which effectively converts a list ls
to an IntSet
, a balanced BST:
def toTree(ls: List[Int]): IntSet = { | |
def toTreeAux(ls: List[Int], n: Int): (List[Int], IntSet) = { | |
if (n <= 0) | |
(ls, Empty) | |
else { | |
val (lls, lt) = toTreeAux(ls, n / 2) // construct left sub-tree | |
val x :: xs = lls // extract root node | |
val (xr, rt) = toTreeAux(xs, n - n / 2 - 1) // construct right sub-tree | |
(xr, IntSet(x, lt, rt)) // construct tree | |
} | |
} | |
val (ls_1, tree) = toTreeAux(ls, List.length(ls)) | |
tree | |
} |
Any comment or query regarding this post is highly appreciated. Thanks.