Like I said before, figuring out when to use parens is the tricky part. :) MeowChow pointed out that my solution didn't account for operators which aren't associative ("2 3 4 + -" should be "2 - (3 + 4)" instead of "2 - 3 + 4"). Then I noted that this information was missing from the data provided in the original problem.

I decided to tackle the problem of associativity, and came up with this solution, at 171 characters in the body of the sub:

%o = ('+' => 1,
      '-' => 1,
      '*' => 2,
      '/' => 2,
     );

%a = ('-' => 1,
      '/' => 1,
     );

sub rpn2ltr {
for$i(@r=@_){if($p=$o{$i}){$r=$p>($r=pop)->[0]||($p==$r->
[0]&&$a{$i})?"($r->[1])":$r->[1];$l=$p>($l=pop)->[0]?"($l->[
1])":$l->[1]}push@_,[$p||9,$p?"$l $i $r":$i]}$_[-1][1]
}
[download]

It's pretty ugly. :/ I hope someone will provide a more elegant solution!