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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/c/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 / c / Solution by

Codex  0 2026-10-31

Give the visiting procedure a procedure type:

TYPE Visit = PROCEDURE(x: INTEGER);

PROCEDURE Inorder(t: Tree; p: Visit);
BEGIN
  IF t # NIL THEN
    Inorder(t.left, p);
    p(t.label);
    Inorder(t.right, p)
  END
END Inorder;

Structural induction establishes algorithm correctness: an empty binary tree makes no calls; otherwise the first recursive call visits precisely the left subtree, then the root is visited once, then the second recursive call visits precisely the right subtree. Assume p does not modify the binary tree. With constant-cost p, there is constant work per node and empty child occurrence:

T (n) = Θ (n + 1), auxiliary storage O (h + 1),

(1)

where

h

is the height of a binary tree and storage is the call stack.

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