As pointed out in the comments, since each element is indeed touched only once, the time complexity is intuitively O(N).
However, because each recursive call to flatten creates a new intermediate array, the run-time depends strongly on the structure of the input array.
A non-trivial1 example of such a case would be when the array is organized similarly to a full binary tree:
[[[a, b], [c, d]], [[e, f], [g, h]]], [[[i, j], [k, l]], [[m, n], [o, p]]]
|
______ + ______
| |
__ + __ __ + __
| | | |
_ + _ _ + _ _ + _ _ + _
| | | | | | | | | | | | | | | |
a b c d e f g h i j k l m n o p
The time complexity recurrence relation is:
T(n) = 2 * T(n / 2) + O(n)
Where 2 * T(n / 2) comes from recursive calls to flatten the sub-trees, and O(n) from pushing2 the results, which are two arrays of length n / 2.
The Master theorem states that in this case T(N) = O(N log N), not O(N) as expected.
1) non-trivial means that no element is wrapped unnecessarily, e.g. [[[a]]].
2) This implicitly assumes that k push operations are O(k) amortized, which is not guaranteed by the standard, but is still true for most implementations.
A "true" O(N) solution will directly append to the final output array instead of creating intermediate arrays:
function flatten_linear(items) {
const flat = [];
// do not call the whole function recursively
// ... that's this mule function's job
function inner(input) {
if (Array.isArray(input))
input.forEach(inner);
else
flat.push(input);
}
// call on the "root" array
inner(items);
return flat;
}
The recurrence becomes T(n) = 2 * T(n / 2) + O(1) for the previous example, which is linear.
Again this assumes both 1) and 2).