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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/imperative-programming-resit/1/d/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Imperative programming (resit) / 1 / d / Solution by

Codex  0 2026-10-31

Split the half-open array segment at its midpoint, using one entry for the root:

PROCEDURE MakeTree(VAR u: ARRAY OF INTEGER;
                   a, b: INTEGER): Tree;
VAR m: INTEGER; t: Tree;
BEGIN
  IF a = b THEN RETURN NIL END;
  m := a + (b-a) DIV 2;
  NEW(t);
  t.label := u[m];
  t.left := MakeTree(u, a, m);
  t.right := MakeTree(u, m+1, b);
  RETURN t
END MakeTree;

The precondition is

0 \leq a \leq b \leq LEN (u)

. Set

n = b - a

. For

n > 0

the child sizes are

⌊ n /2 ⌋

and

n - 1 - ⌊ n /2 ⌋

, differing by at most one. The same construction in both subtrees proves subtree-size balance at every node. Structural induction also gives the inorder traversal

u [a .. m)

, followed by

u [m]

, followed by

u [m + 1.. b)

, hence precisely

u [a .. b)

; no sortedness assumption is needed.

Exactly

n

records are allocated. Each nonempty recursive call makes two child calls, so there are

2 n + 1

calls including empty segments. Under constant-cost allocation and reference parameter passing,

T (n) = Θ (n + 1), inorder (t) = u [a .. b), t is nearly balanced .

(1)

For positive

n

this is proportional to

b - a

; the empty case takes constant time. The call stack has

O (1 + lo g (n + 1))

depth.

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