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
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 .
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:
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 .