This problem is a little bit hard to think at first. But when you write down for a few samples, you wiil discover the patterns inside them, let us get started:
if n = 0: 0 -- 0
if n = 1: 0 -- 0, 1 -- 1
if n = 2: 00 -- 0, 01 -- 1, 11 -- 3, 10 -- 2
if n = 3: 000 -- 0, 001 -- 1, 011 -- 3, 010 -- 2, 110 -- 6, 111 -- 7, 101 -- 5, 100 -- 4.
Did you find out the pattern, we denote the result as f(n), it is just like:
f(n) = [0, f(n-1)] + [1, reverse(f(n-1))]