# Problem 12.3

## Average node depth in a randomly built binary search tree

In this problem, we prove that the average depth of a node in a randomly built binary search tree with $n$ nodes is $\O(\lg n)$. Although this result is weaker than that of Theorem 12.4, the technique we shall use reveals a surprising similarity between the building of a binary search tree and the execution of

`RANDOMIZED-QUICKSORT`

from Section 7.3.We define the

total path length$P(T)$ of a binary search tree as the sum, over all nodes $x$ in $T$, of the depth of node $x$, which we denote by $d(x, T)$.

- Argue that the average depth of a node in $T$ is $$ \frac{1}{n}\sum_{x \in T}^{n-1} d(x, T) = \frac{1}{n} P(T) $$
Thus, we wish to show that the expected value of $P(T)$ is $O(n \lg n)$.

- Let $T_L$ and $T_R$ denote the left and right subtrees of tree $T$, respectively. Argue that if $T$ has $n$ nodes, then
$$ P(T) = P(T_L) + P(T_R) + n - 1 $$

- Let $P(n)$ denote the average total path length of a randomly built binary search tree with $n$ nodes. Show that:
$$ P(n) = \frac{1}{n} \sum_{i=0}^{n-1}(P(i) + P(n-i-1) + n - 1) $$

- Show how to rewrite $P(n)$ as $$ P(n) = \frac{2}{n} \sum_{k=1}^{n-1} P(k) + \Theta(n) $$
- Recalling the alternative analysis of the randomized version of quicksort given in Problem 7.3, conclude that $P(n) = \O(n \lg n)$.
At each recursive invocation of quicksort, we choose a random pivot element to partition the set of elements being sorted. Each node of a binary search tree partitions the set of elements that fall into the subtree rooted at that node.

- Describe an implementation of quicksort in which the comparisons to sort a set of elements are exactly the same as the comparisons to insert the elements into a binary search tree. (The order in which comparisons are made may differ, but the same comparisons must occur.)

### Average depth of the node

This is literally by definition.

### Total path length expressed through subtrees

Let's say that $T_L$ contains $l$ nodes. If we add a new node above $T_L$ as a new root of the tree, each of the $l$ nodes will get one extra edge connecting it to the root. A similar argument can be made for $T_R$ with $r$ nodes. Thus, when we take two trees, $T_L$ and $T_R$ and make them the left and right subtrees of a new tree, we add $l + r$ to the total path length of both trees, as each node now has an extra edge to the root. Observe that if the new tree has $n$ elements, then $n - 1 = l + r$. Hence we have that:

$$ P(T) = P(T_L) + P(T_R) + n - 1 $$

### Average total path length of randomly built tree

When we're building the tree randomly, each element of $\{1, 2, \ldots, n\}$ is equally likely to be the root. This follows pretty straightforwardly from the last two parts.

### Rewritten expression

We need to observe that each $P(i)$ appears twice because of the sum. That is:

$$ \begin{aligned} P(n) &= \frac{1}{n} \sum_{i=0}^{n-1}(P(i) + P(n-i-1) + n - 1) \\ &= \frac{1}{n} \sum_{i=0}^{n-1}(P(i) + P(n-i-1)) + \frac{1}{n} \sum_{i=0}^{n-1}(n - 1) \\ &= \frac{1}{n} \Big( P(0) + P(n - 1) + P(1) + P(n - 2) + \ldots + P(n - 2) + P(1) + P(n - 1) + P(0) \Big) + \Theta(n) \\ &= \frac{2}{n} \sum_{k=0}^{n-1} P(k) \\ &= \frac{2}{n} \sum_{k=1}^{n-1} P(k) \\ \end{aligned} $$

The last step holds because $P(0) = 0$.

### Upper bound

Problem 7.3 established that:

$$ \sum_{k=2}^{n-1}k\lg{k} \le \frac{1}{2}n^2\lg{n} - \frac{1}{8}n^2 $$

Keeping that in mind, let's use the substitution method to establish an upper bound on $P(n)$. Let's assume $O(n\lg{n})$ and plug in $an\lg{n} + b$.

$$ \begin{aligned} P(n) &= \frac{2}{n} \sum_{k=1}^{n-1}P(k) + \Theta(n) \\ &= \frac{2}{n} \sum_{k=2}^{n-1}P(k) + \Theta(n) && \text{(because }P(1) = 0 \text{)} \\ &\le \frac{2}{n} \sum_{k=2}^{n-1}(ak\lg{k} + b) + \Theta(n) \\ &= \frac{2a}{n} \sum_{k=2}^{n-1}(k\lg{k}) + \frac{2b}{n}(n - 1) + \Theta(n) \\ &\le \frac{2a}{n} \sum_{k=2}^{n-1}(k\lg{k}) + 2b + \Theta(n) \\ &\le \frac{2a}{n} \left( \frac{1}{2} n^2 \lg{n} - \frac{1}{8} n^2 \right) + 2b + \Theta(n) \\ &= an \lg{n} - \frac{4}{a} n + 2b + \Theta(n) \\ &= an \lg{n} + b + \left(\Theta(n) + b - \frac{4}{a}n\right) \\ &\le an \lg{n} + b \end{aligned} $$

If we choose a large enough $a$ so $\frac{a}{4}n \ge \Theta(n) + b$. Chapter 4 advises against asymptotic notation when using the substitution method, but I like to live dangerously.

### Describe an implementation of quicksort(?)

In quicksort, once a pivot is chosen, every element is compared against it. Similarly, once the root is chosen, every other element is compared against it. Applying this thinking recursively, we end up at the same number of comparisons.