SKEDSOFT

Dynamic order statistics: Data structure-such as a doubly linked list, a hash table, or a binary search tree-but many others require a dash of creativity. Only in rare situations will you need to create an entirely new type of data structure, though. More often, it will suffice to augment a textbook data structure by storing additional information in it.

Specifically, the ith order statistic of a set of n elements, where i ∈ {1, 2,..., n}, is simply the element in the set with the ith smallest key. We saw that any order statistic could be retrieved in O(n) time from an unordered set. In this section, we shall see how red-black trees can be modified so that any order statistic can be determined in O(lg n) time. We shall also see how the rank of an element-its position in the linear order of the set-can likewise be determined in O(lg n) time.

A data structure that can support fast order-statistic operations is shown in Figure 14.1. An order-statistic tree T is simply a red-black tree with additional information stored in each node. Besides the usual red-black tree fields key[x], color[x], p[x], left[x], and right[x] in a node x, we have another field size[x]. This field contains the number of (internal) nodes in the subtree rooted at x (including x itself), that is, the size of the subtree. If we define the sentinel's size to be 0, that is, we set size[nil[T]] to be 0, then we have the identity

size[x] = size[left[x]] size[right[x]] 1.

We do not require keys to be distinct in an order-statistic tree. (For example, the tree in Figure 14.1 has two keys with value 14 and two keys with value 21.) In the presence of equal keys, the above notion of rank is not well defined. We remove this ambiguity for an order-statistic tree by defining the rank of an element as the position at which it would be printed in an inorder walk of the tree. In Figure 14.1, for example, the key 14 stored in a black node has rank 5, and the key 14 stored in a red node has rank 6.

Retrieving an element with a given rank

Before we show how to maintain this size information during insertion and deletion, let us examine the implementation of two order-statistic queries that use this additional information. We begin with an operation that retrieves an element with a given rank. The procedure OS-SELECT(x, i) returns a pointer to the node containing the ith smallest key in the subtree rooted at x. To find the ith smallest key in an order-statistic tree T , we call OS-SELECT(root[T], i).

	OS-SELECT(x, i)
1  r ← size[left[x]] 1
2  if i = r
3     then return x
4  elseif i< r
5     then return OS-SELECT(left[x], i)
6  else return OS-SELECT(right[x], i - r)

The value of size[left[x]] is the number of nodes that come before x in an inorder tree walk of the subtree rooted at x. Thus, size[left[x]] 1 is the rank of x within the subtree rooted at x.

In line 1 of OS-SELECT, we compute r, the rank of node x within the subtree rooted at x. If i = r, then node x is the ith smallest element, so we return x in line 3. If i< r, then the ith smallest element is in x's left subtree, so we recurse on left[x] in line 5. If i > r, then the ith smallest element is in x's right subtree. Since there are r elements in the subtree rooted at x that come before x's right subtree in an inorder tree walk, the ith smallest element in the subtree rooted at x is the (i - r)th smallest element in the subtree rooted at right[x]. This element is determined recursively in line 6.

To see how OS-SELECT operates, consider a search for the 17th smallest element in the order-statistic tree of Figure 14.1. We begin with x as the root, whose key is 26, and with i = 17. Since the size of 26's left subtree is 12, its rank is 13. Thus, we know that the node with rank 17 is the 17 - 13 = 4th smallest element in 26's right subtree. After the recursive call, x is the node with key 41, and i = 4. Since the size of 41's left subtree is 5, its rank within its subtree is 6. Thus, we know that the node with rank 4 is the 4th smallest element in 41's left subtree. After the recursive call, x is the node with key 30, and its rank within its subtree is 2. Thus, we recurse once again to find the 4 × 2 = 2nd smallest element in the subtree rooted at the node with key 38. We now find that its left subtree has size 1, which means it is the second smallest element. Thus, a pointer to the node with key 38 is returned by the procedure.

Because each recursive call goes down one level in the order-statistic tree, the total time for OS-SELECT is at worst proportional to the height of the tree. Since the tree is a red-black tree, its height is O(lg n), where n is the number of nodes. Thus, the running time of OS-SELECT is O(lg n) for a dynamic set of n elements.