The recurrence T(n) = 2T(n/2) + 2 with T(1) = c solves by iteration to

Unrolling k = log₂ n levels: T(n) = 2^k · T(1) + 2(2^k - 1). With 2^k = n, this is O(n). The min-max divide-and-conquer algorithm hits exactly this recurrence the work doubles at each level but the per-level constant is small so the dominant term is the bottom recursion which is O(n).