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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/imperative-programming-resit/3/c/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Imperative programming (resit) / 3 / c / Solution by

Codex  0 2026-10-31

Perform a depth-first traversal, retaining the path's letters in a buffer. A terminal node must be printed before its children, so a word precedes its proper extensions in lexicographic order.

VAR word: ARRAY N OF CHAR;

PROCEDURE Visit(t: Trie; depth: INTEGER);
VAR j, c: INTEGER;
BEGIN
  IF t # NIL THEN
    IF t.terminal THEN
      j := 0;
      WHILE j < depth DO
        Out.Char(word[j]); j := j+1
      END;
      Out.Ln
    END;
    c := 0;
    WHILE c < 26 DO
      IF t.child[c] # NIL THEN
        word[depth] := CHR(ORD("a")+c);
        Visit(t.child[c], depth+1)
      END;
      c := c+1
    END
  END
END Visit;

PROCEDURE PrintTrie(t: Trie);
BEGIN
  Visit(t, 0)
END PrintTrie;

The recursion invariant is that word[0..depth) spells the path to t. The child loop's loop invariant says children with indices below c have been completely printed. Each edge appends its lowercase label; each terminal flag prints exactly one represented string. All words therefore appear once, in lowercase. The output loop prints precisely the prefix of length depth; no terminating zero needs storing in this buffer.

An N-character buffer suffices for words of length at most N. At depth N there can be no child, so word[N] is never accessed. A terminal root prints an empty line; a NIL root prints nothing. The running time is linear in allocated nodes plus the number of printed characters, since the alphabet size is fixed at 26.

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