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: A Randomized Binary Search Tree
The problem with random binary search trees is, of course, that they are
not dynamic. They don't support the
or
operations
needed to implement the
interface. In this section we describe
a data structure called a
that uses Lemma 7.1 to implement
the
interface.7.2
A node in a
is like a node in a
in that it has
a data value,
, but it also contains a unique numerical priority,
, that is assigned at random:
class TreapNode : public BSTNode<Node, T> {
friend class Treap<Node,T>;
int p;
};
In addition to being a binary search tree, the nodes in a
also obey the heap property:
, except the root,
.
|
The heap and binary search tree conditions together ensure that, once
the key (
) and priority (
) for each node are defined, the
shape of the
is completely determined. The heap property tells us that
the node with minimum priority has to be the root,
, of the
.
The binary search tree property tells us that all nodes with keys smaller
than
are stored in the subtree rooted at
and all nodes
with keys larger than
are stored in the subtree rooted at
.
The important point about the priority values in a
is that they
are unique and assigned at random. Because of this, there are
two equivalent ways we can think about a
. As defined above, a
obeys the heap and binary search tree properties. Alternatively,
we can think of a
as a
whose nodes
were added in increasing order of priority. For example, the
in Figure 7.5 can be obtained by adding the sequence of
values
.
Since the priorities are chosen randomly, this is equivalent to taking a random permutation of the keys -- in this case the permutation is
. But this means that the
shape of a treap is identical to that of a random binary search tree.
In particular, if we replace each key
by its rank,7.3 then Lemma 7.1 applies.
Restating Lemma 7.1 in terms of
s, we have:
that stores a set 
of
keys, the following statements hold:
, the expected length of
the search path for
is
.
, the expected length of the
search path for
is
.
denotes the rank of
in the set
.
Lemma 7.2 tells us that
s can implement the
operation efficiently. However, the real benefit of a
is that
it can support the
and
operations. To
do this, it needs to perform rotations in order to maintain the heap property. Refer to Figure 7.6.
A rotation in a binary
search tree is a local modification that takes a parent
of a node
and makes
the parent of
, while preserving the binary search tree
property. Rotations come in two flavours: left or right
depending on whether
is a right or left child of
, respectively.
The code that implements this has to handle these two possibilities
and be careful of a boundary
case (when
is the root) so the actual code is a little longer than
Figure 7.6 would lead a reader to believe:
void rotateLeft(Node *u) {
Node *w = u->right;
w->parent = u->parent;
if (w->parent != nil) {
if (w->parent->left == u) {
w->parent->left = w;
} else {
w->parent->right = w;
}
}
u->right = w->left;
if (u->right != nil) {
u->right->parent = u;
}
u->parent = w;
w->left = u;
if (u == r) { r = w; r->parent = nil; }
}
void rotateRight(Node *u) {
Node *w = u->left;
w->parent = u->parent;
if (w->parent != nil) {
if (w->parent->left == u) {
w->parent->left = w;
} else {
w->parent->right = w;
}
}
u->left = w->right;
if (u->left != nil) {
u->left->parent = u;
}
u->parent = w;
w->right = u;
if (u == r) { r = w; r->parent = nil; }
}
In terms of the
data structure, the most important property of a
rotation is that the depth of
decreases by one while the depth of
increases by one.
Using rotations, we can implement the
operation as follows:
We create a new node,
, and assign
and pick a random value
for
. Next we add
using the usual
algorithm
for a
, so that
is now a leaf of the
.
At this point, our
satisfies the binary search tree property,
but not necessarily the heap property. In particular, it may be the
case that
. If this is the case, then we perform a
rotation at node
=
so that
becomes the parent of
.
If
continues to violate the heap property, we will have to repeat this, decreasing
's depth by one every time, until
either becomes the root or
.
bool add(T x) {
Node *u = new Node;
u->x = x;
u->p = rand();
if (BinarySearchTree<Node,T>::add(u)) {
bubbleUp(u);
return true;
}
return false;
}
void bubbleUp(Node *u) {
while (u->parent != nil && u->parent->p > u->p) {
if (u->parent->right == u) {
rotateLeft(u->parent);
} else {
rotateRight(u->parent);
}
}
if (u->parent == nil) {
r = u;
}
}
An example of an
operation is shown in Figure 7.7.
The running time of the
operation is given by the time it
takes to follow the search path for
plus the number of rotations
performed to move the newly-added node,
, up to its correct location
in the
. By Lemma 7.2, the expected length of the
search path is at most
. Furthermore, each rotation
decreases the depth of
. This stops if
becomes the root, so
the expected number of rotations cannot exceed the expected length of
the search path. Therefore, the expected running time of the
operation in a
is
. (Exercise 7.5
asks you to show that the expected number of rotations performed during
an addition is actually only 
.)
The
operation in a
is the opposite of the
operation. We search for the node,
, containing
and then perform
rotations to move
downwards until it becomes a leaf and then we
splice
from the
. Notice that, to move
downwards, we can
perform either a left or right rotation at
, which will replace
with
or
, respectively.
The choice is made by the first of the following that apply:
and
are both
, then
is a leaf and no rotation is performed.
(or
) is
, then perform a right (or left, respectively) rotation at
.
(or
, then perform a right rotation (or left rotation, respectively) at
.
doesn't become disconnected and that the heap property is restored once
is removed.
bool remove(T x) {
Node *u = findLast(x);
if (u != nil && compare(u->x, x) == 0) {
trickleDown(u);
splice(u);
delete u;
return true;
}
return false;
}
void trickleDown(Node *u) {
while (u->left != nil || u->right != nil) {
if (u->left == nil) {
rotateLeft(u);
} else if (u->right == nil) {
rotateRight(u);
} else if (u->left->p < u->right->p) {
rotateRight(u);
} else {
rotateLeft(u);
}
if (r == u) {
r = u->parent;
}
}
}
An example of the
operation is shown in Figure 7.8.
The trick to analyze the running time of the
operation is
to notice that this operation is the reverse of the
operation.
In particular, if we were to reinsert
, using the same priority
,
then the
operation would do exactly the same number of rotations
and would restore the
to exactly the same state it was in before
the
operation took place. (Reading from bottom-to-top,
Figure 7.8 illustrates the addition of the value 9 into a
.) This means that the expected running time of the
on a
of size
is proportional to the expected running time
of the
operation on a
of size
. We conclude
that the expected running time of
is
.
The following theorem summarizes the performance of the
data
structure:
implements the
interface. A
supports
the operations
,
, and
in
expected time per operation.
It is worth comparing the
data structure to the
data structure. Both implement the
operations in
expected time per operation. In both data structures,
and
involve a search and then a constant number of pointer changes
(see Exercise 7.5 below). Thus, for both these
structures, the expected length of the search path is the critical value
in assessing their performance. In a
, the expected length
of a search path is
, the expected length of a search path is
are considerably shorter and this
translates into noticeably faster operations on
s than
s.
Exercise 4.7 in Chapter 4 shows how the
expected length of the search path in a
can be reduced to
is noticeably longer than in
a
.
comes from the fact
that this data structure is simultaneously a binary search tree
(Section 6.2) and a heap (Chapter 10).
in a set of elements is the number of
elements in that are less than
.
