Let's first see in general how addition works in any base system.
Let A and B be two numbers, for any position 'i' in A and B
$Sum_{i} = (a_{i} + b_{i} + c_{i-1})$%$Base$, where $c_{i-1}$ represents carry from previous position.
$Carry_{i} = (Sum_{i} )/Base$.
So, this formula can be used to add given n-bit binary number, where base will be 2.
Time Complexity = $\mathcal{O}(N)$
Space Complexity = $\mathcal{O}(N)$, for storing the result.
n = A.length
carry = 0
for i = n to 1
C[i] = (A[i] + B[i] + carry) % 2
carry = (A[i] + B[i] + carry) / 2
C[i] = carry
return C