Answer:
int ComputeFibonacci(int N) {
if(N == 0)
return 0;
else if (N == 1)
return 1;
else
return ComputeFibonacci(N-1) + ComputeFibonacci(N-2);
}
Explanation:
Inside the function ComputeFibonacci that takes one parameter, N, check the base cases first. If N is eqaul to 0, return 0. If N is eqaul to 1, return 1. Otherwise, call the ComputeFibonacci function with parameter N-1 and N-2 and sum these and return the result.
For example,
If N = 4 as in the main part:
ComputeFibonacci(4) → ComputeFibonacci(3) + ComputeFibonacci(2) = 2 + 1 = 3
ComputeFibonacci(3) → ComputeFibonacci(2) + ComputeFibonacci(1) = 1 + 1 = 2
ComputeFibonacci(2) → ComputeFibonacci(1) + ComputeFibonacci(0) = 1 + 0 = 1
*Note that you need to insert values from the bottom. Insert the values for ComputeFibonacci(1) and ComputeFibonacci(0) to find ComputeFibonacci(2) and repeat the process.
