Recursion for "Sum All Odd Fibonaccis" Challenge

JavaScript
1

Hi together
It took me a few days, but I passed the “Sum All Odd Fibonaccis” Challenge.

I have 2 Questions:
1 How good/bad is my code (I think not to bad)?
2 Actually it should be possible to get all Fibonacci Numbers by using Recursion. But can I use Recursion to solve this challenge too? Including just sum the odd Numbers?

Your code so far

function sumFibs(num) {
let fibs = [1 ,1], temp = fibs[fibs.length - 1];

while (temp <= num) {
  temp += fibs[fibs.length - 2];
  if (temp <= num) fibs.push(temp);
}

return fibs.reduce((accumulator, current) => { 
  return ((current % 2) !== 0) ? (accumulator + current) : (accumulator + 0);
});
}

sumFibs(4);

Your browser information:

User Agent is: Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:85.0) Gecko/20100101 Firefox/85.0.

Challenge: Sum All Odd Fibonacci Numbers

Link to the challenge:

2

Here’s one example with a recursive solution. The function takes the latest two Fibonacci numbers (n and m) and the sum of all odd numbers as parameters.

function sumFibs(num) {

  const fibsumRecursion = (n, m, sum) => {

    if (m <= num ){                   /* as long as the latest number is smaller than num... */

      sum = m%2 !== 0 ? sum+m : sum;  /* if the  latest number is odd, add it to sum */
      [n,m] = [m, n+m];               /* use array destructuring to find the next two numbers */
      return fibsum(n, m, sum)        /* run the function again with new parameters */

    } else {
      return sum
    } 
  }
  return fibsumRecursion(1,1,1)       /* initialise with n=1, m=1 --> sum is already 1*/
}

I’m not sure if this is super readable, but it avoids building the whole array of numbers first.