2 of 3 added 248 characters in body

Python 2: 59 chars

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.

The specs don't allow this, but if one could write a function instead of a program and output the empty string for 0, a 40 char solution would be possible.

g=lambda n:n and g(-~n/3)+"0+-"[n%3]or""