Python 2: 60 chars
---------------

<!-- language: lang-python -->

    n=input()
    s=""
    while n:s="0+-"[n%3]+s;n=-~n/3;
    print s or'0'

Generates the balanced ternary digit-by-digit from the end. The last digit is given by the residue `n%3` being `-1`,`0`, or `+1`. We then remove the last digit and divide by 3 using Python's floor-divide `n=(n+1)/3`. Then, we proceed recursively with the new last digit until the number is 0.

A special case is needed for the input `0` to give `0` rather than the empty string.