Complexity of finding 7th smallest element in a min-heap? I don't know if it's allowed to traverse nodes in a min-heap.

Because, if traversal is allowed then for finding $7^{th}$ smallest element, only a constant number of nodes need to be checked, thus giving the complexity of $\Theta(1)$.

But if traversing is not allowed, then for finding $7^{th}$ smallest element, I'll need to call extract_min() 6 times, leading to a complexity of $\Theta(\log n)$, which I think should be correct, but not sure. 

My understanding is, if I'm traversing, then it's not a min-heap, it will be an augmented min-heap. Is it correct?
If you had a "black-box" min heap where the only operations available to you are adding elements and extracting (reading and simultaneously removing) the smallest item, then the time complexity is O (n) obviously. 

It is imaginable that you have a specific implementation where even finding the second smallest element cannot be done in constant time. You might have in your data structure O (log n) items that might be the second smallest, and the O (log n) operation to remove the smallest item will also determine the second largest. I find it hard to describe an algorithm that works that way, but I'm sure it's possible. 

The way a min-heap is usually implemented (or the way it is implemented if I had to implement it, or the way it is implemented in a random text book), finding the 7th smallest element can be done in constant time.