从AVL树获得中位数？

Question

If you have an AVL tree, what's the best way to get the median from it? The median would be defined as the element with index ceil(n/2) (index starts with 1) in the sorted list.

So if the list was

1 3 5 7 8

the median is 5. If the list was

1 3 5 7 8 10

the median is 5.

If you can augment the tree, I think it's best to let each node know the size (number of nodes) of the subtree, (ie 1 + left.size + right.size). Using this, the best way I can think of makes median searching O(lg n) time because you can traverse by comparing indexes.

Is there a better way?

Answer 1

Augmenting the AVL tree to store subtree sizes is generally the best approach here if you need to optimize over median queries. It takes time O(log n), which is pretty fast.

If you'll be computing the median a huge number of times, you could potentially use an augmented tree and also cache the median value so that you can read it in time O(1). Each time you do an insertion or deletion, you might need to recompute the median in time O(log n), which will slow things down a bit but not impact the asymptotic costs.

Another option would be to thread a doubly-linked list through the nodes in the tree so that you can navigate from a node to its successor or predecessor in constant time. If you do that, then you can store a pointer to the median element, and then on an insertion or a deletion, move the pointer to the left or to the right as appropriate. If you delete the median itself, you can just move the pointer left or right as you'd like. This doesn't require any augmentation and might be a bit faster, but it adds two extra pointers into each node.

Hope this helps!