In mathematics, the Fibonacci numbers are the numbers in the following integer sequence, called the Fibonacci sequence, and characterized by the fact that every number after the first two is the sum of the two preceding ones:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ...
Time Complexity for generate fibonacci series using recursion is O(2^n).
To know more about calculating time complexity please do refer fibonacci-series-complexity
Let's generate a fibonacci series upto 5 numbers. i.e., fibonacciNumbers(5), where n = 5
if (n (5) === 1) --> False
else
if (n (5) === 2) --> False
else
s = fibonacciNumbers(n(5) - 1) = fibonacciNumbers(4)
(s = [0,1,1,2]) ^ |
(s.push(s[3](2) + s[2](1)) = 3)) | |
return s ([0,1,1,2,3]) | |
| | V
| | if (n (4) === 1) --> False
| | else
| | if (n (4) === 2) --> False
| | else
| | s = fibonacciNumbers(n(4) - 1) = fibonacciNumbers(3)
| | (s=[0,1,1]) ^ |
| | s.push(s[2](1) + s[1](1) = 2) | |
| --* return s ([0,1,1,2]) | V
| | if (n (3) === 1) --> False
| | else
| | if (n (3) === 2) --> False
| |
| | else
| | s = fibonacciNumbers(n(3) - 1) = fibonacciNumbers(2) <-----------
| | (s=[0,1]) | |
| | s.push(s[1](1) + s[0](0) = 1). | |
| --* return s ([0,1,1]) | |
| | |
| V |
| if (n (2) === 1)-->False |
| else |
V if (n (2) === 2)-->True |
fibonacciNumbers(5) = [0,1,1,2,3]; return [0,1] *------------