Bijective mapping from integers to a variable number of bits

Python, 20 bytes

lambda n:bin(~n)[4:]

Test:

>> [bin(~n)[4:] for n in range(16)]
['', '0', '1', '00', '01', '10', '11', '000', '001', '010', '011', '100', '101', '110', '111', '0000']

Doing lambda n:bin(n+1)[3:] increments the input, then takes the binary representation with the first symbol removed ([3:] because the prefix 0b is two chars chars). Since any positive number starts with 1 in binary, this uniquely gives a binary representation.

A byte is saved by instead using the bit complement ~n to get the negation -(n+1), and removing the negative sign by chopping off one more symbol.

Jelly, 3 bytes

‘BḊ

Same idea as xnor: maps 0 1 2 3 4 ... to [] [0] [1] [0 0] [0 1] ...; the code is basically increment → binary → remove first.

Try it out online.

Pyth, 5 bytes

t.BhQ

Simply a translation of xnor's answer into Pyth.

Q is eval()'d input(), h increments it, .B converts it to a binary string, and t takes the "tail" (which is everything except the first character).