Consider the two binary search trees shown in Figure 7.1, each of
which has
nodes. The one on the left is a list and the other
is a perfectly balanced binary search tree. The one on the left has a
height of
and the one on the right has a height of three.
Imagine how these two trees could have been constructed. The one on the left occurs if we start with an empty BinarySearchTree and add the sequence
The above example gives some anecdotal evidence that, if we choose a
random permutation of
, and add it into a binary search
tree, then we are more likely to get a very balanced tree (the right
side of Figure 7.1) than we are to get a very unbalanced tree
(the left side of Figure 7.1).
We can formalize this notion by studying random binary search trees.
A random binary search tree
of size
is obtained in the
following way: Take a random permutation,
,
of the integers
and add its elements, one by one,
into a BinarySearchTree. By random permutation
we mean that
each of the possible
permutations (orderings) of
is equally likely, so that the probability of obtaining any particular
permutation is
.
Note that the values
could be replaced by any ordered
set of
elements without changing any of the properties of the
random binary search tree. The element
is
simply standing in for the element of rank
in an ordered set of
size
.
Before we can present our main result about random binary search trees,
we must take some time for a short digression to discuss a type of number
that comes up frequently when studying randomized structures. For a
non-negative integer, , the
-th harmonic number,
denoted
, is defined as
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We will prove Lemma 7.1 in the next section. For now, consider what
the two parts of Lemma 7.1 tell us. The first part tells us that if
we search for an element in a tree of size
, then the expected length
of the search path is at most
. The second part tells
us the same thing about searching for a value not stored in the tree.
When we compare the two parts of the lemma, we see that it is only
slightly faster to search for something that is in the tree compared to
something that is not.
The key observation needed to prove Lemma 7.1 is the following:
The search path for a value
in the open interval
in a
random binary search tree,
, contains the node with key
if, and only if, in the random permutation used to create
,
appears before any of
.
To see this, refer to Figure 7.3 and notice that until
some value in
is added, the search
paths for each value in the open interval
are identical. (Remember that for two values to have
different search paths, there must be some element in the tree that
compares differently with them.) Let
be the first element in
to appear in the random permutation.
Notice that
is now and will always be on the search path for
.
If
then the node
containing
is created before the
node
that contains
. Later, when
is added, it will be
added to the subtree rooted at
, since
. On the other
hand, the search path for
will never visit this subtree because it
will proceed to
after visiting
.
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Similarly, for
,
appears in the search path for
if and only if
appears before any of
in the random permutation used to
create
.
Notice that, if we start with a random permutation of
,
then the subsequences containing only
and
are also random
permutations of their respective elements. Each element, then, in the
subsets
and
is equally likely to appear before
any other in its subset in the random permutation used to create
.
So we have
With this observation, the proof of Lemma 7.1 involves some simple calculations with harmonic numbers:
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The following theorem summarizes the performance of a random binary search tree:
We should emphasize again that the expectation in Theorem 7.1 is with
respect to the random permutation used to create the random binary
search tree. In particular, it does not depend on a random choice of
; it is true for every value of
.