Solution

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

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Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Imperative programming / 1 / e / Solution by Codex 0 2026-10-31
Maintain a cursor to the first unread marker. A fork consumes its own marker and then two complete encodings, in left-to-right order.
PROCEDURE Decode(VAR code: ARRAY OF INTEGER;
                 VAR pos: INTEGER): Tree;
VAR marker: INTEGER; left, right: Tree;
BEGIN
  marker := code[pos]; INC(pos);
  IF marker = 0 THEN RETURN 0 END;
  ASSERT(marker = 1);
  left := Decode(code, pos);
  right := Decode(code, pos);
  RETURN NewFork(left, right)
END Decode;
Start with pos := 0; t := Decode(code, pos), with sufficient unused arena space. The valid-input precondition ensures that every read exists and every fork has two encoded subtrees. Evaluating the two recursive calls in separate statements fixes their order.
Structural induction shows that decoding reconstructs the same binary tree shape and leaves the cursor just after its encoding. The tree handles need not coincide with the original handles: the abstract tree does. The result is
(1)
The time complexity is linear in the encoding length, with call stack space bounded by the height of a binary tree.
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