Hello!
Do the following exercise:
count(3);
n = 3;
n === 1? false // return [1] skipped.
// Here count is called again, hence this steps is paused.
var numbers = count(n - 1); // n would be 2
# STEP1
n = 2;
n === 1? false // return [1] skipped.
// Here count is called again, hence this steps is paused.
var numbers = count(n - 1); // n would be 1
// Here the function is paused while the count(n-1) is executed
#STEP2
n = 1;
n === 1? true // an array with one is returned.
// Now the function call retraces its steps (yep, backwards now)
#Continue with STEP2
// Remember, n = 2
numbers = [1];
numbers.push(2);
return [1, 2];
#Continue with STEP1
// Remember, n = 3
numbers = [1, 2];
numbers.push(3);
return [1, 2, 3];
It helps me understand some algorithms, I hope it helps You too 
.