Indeed. If the smallest element is deleted, then we are removing the singleton element of the first heap. We need to find the second-smallest element, which will be in the next heap after it, but it could be any of the leaves. That means we need to traverse O(heapsize) elements.<p>And since we started with the smallest element, this will cascade through each of the heaps, in order to keep them all (but the last) at full size. All together, that&#x27;s O(N).