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My guess is roughly around $2^{d-1}$, but I am not sure how to show it.

Simply put, let $S$ be the set of all rooted ordered unlabelled binary trees, with maximum depth $d$. What is the expected number of nodes/elements in a tree which is chosen uniformly at random from this set $S$.

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    $\begingroup$ Two things need to be clarified here. a) How do you define the "maximum depth" of a tree? I suspect that you might mean rooted trees, in which case there's a standard definition of the maximum depth. Or is the depth to be maximized over all possible roots? b) You haven't defined a distribution for which to answer this question. Are you assuming all possible trees with the given maximum depth to be equiprobable? Are these labeled or unlabeled trees? $\endgroup$ – joriki Jun 30 '18 at 14:02
  • $\begingroup$ Changed to mean rooted trees, Yes I mean all possible trees, which have maximum depth to be exactly $d$ are equiprobable. Unlabelled trees or labelled trees, anything would help. I guess the expected value for both the cases would not be far apart. $\endgroup$ – Vk1 Jun 30 '18 at 14:06
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    $\begingroup$ The answer will differ (I expect, significantly) depending on whether you count all labeled or all unlabeled trees as equiprobale. If you're so indifferent to such a central aspect of the problem, I think it would be helpful to tell us more about the context in which you're considering this problem. (I wrote that before you changed your comment to say that the expected values wouldn't be far apart. I think they will. If you consider all labeled trees to be equiprobable, the ones with the most vertices will dominate, because of the factor $n!$ for the labels.) $\endgroup$ – joriki Jun 30 '18 at 14:09
  • $\begingroup$ Its in context of binary search trees in computer science. If there are $n$ elements in a binary tree, than the expected depth of the tree is $O(\log_2 n)$. I was wondering if the converse is also true. $\endgroup$ – Vk1 Jun 30 '18 at 14:11
  • $\begingroup$ That's not a well-defined "converse". In the case of $n$ elements, you randomize over the possible orders of the elements. In your question, you have a completely different randomization, and hence a completely different result (even if one of the cases does come out near $2^d$). $\endgroup$ – joriki Jun 30 '18 at 14:13
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It's easier to first of all deal with trees of maximum depth up to $d$ and then subtract to get answers for trees of maximum depth exactly $d$.

Let's set up a recurrence relation for the number $a_d$ of rooted ordered unlabeled binary trees of maximum depth up to $d$. It simplifies things to include the empty tree. Then a tree of depth up to $d$ is either the empty tree, or it's a root with each child a tree of depth up to $d-1$. Thus we have the recurrence

$$ a_d = a_{d-1}^2+1\;, $$

with intial value $a_0=1$. This is OEIS A003095. Surprisingly little seems to be known about it. It leads us to OEIS A001699, which is $a_d-a_{d-1}$, and thus the number of rooted ordered unlabeled binary trees of maximum depth exactly $d$. The OEIS entry says that this asymptotically grows as $c^{2^d}$ with an apparently only numerically known constant $c\approx1.5028368$.

Now to get the expected number of elements, we need the sum $s_d$ of the numbers of elements over all trees – again, first the sum for all trees of maximum depth up to $d$ and then, by subtraction, the sum for all tress of maximum depth exactly $d$.

The empty tree contributes no elements to the sum. For the $a_{d-1}^2$ trees with a root, the root contributes $a_{d-1}^2$ to the sum. The sum of the elements of all possible left children of the root is $s_{d-1}$, and they each appear once for each of the $a_{d-1}$ possible right children; and the same vice versa. Thus, the two children contribute a total of $2a_{d-1}s_{d-1}$ to the sum, so overall we get the recurrence

$$ s_d=2a_{d-1}s_{d-1}+a_{d-1}^2\;. $$

Dividing through by $a_{d-1}^2$ and using the fact that for large $d$ we have $a_d=a_{d-1}^2+1\approx a_{d-1}^2$, we arrive at

$$ \frac{s_d}{a_d}\approx2\frac{s_{d-1}}{a_{d-1}}+1\;. $$

The fractions are the mean numbers of elements for trees of maximum depth up to $d$ and $d-1$, respectively. Since $s_d\gg s_{s-1}$ and $a_d\gg a_{d-1}$ for large $d$, this equation also holds approximately when we replace all values for trees of maximum depth up to $d$ by the corresponding values for tree of maximum depth exactly $d$:

$$ \frac{s_d-s_{d-1}}{a_d-a_{d-1}}\approx2\frac{s_{d-1}-s_{d-2}}{a_{d-1}-a_{d-2}}+1\;, $$

and thus

$$ b_d\approx2b_{d-1}+1 $$

for the desired mean number $b_d$ of elements in a tree of maximum depth exactly $d$.

This recurrence has the general solution

$$ b_d\approx\lambda\cdot2^d-1\;. $$

As with the constant $c$ above, since the equation only holds asymptotically for large $d$, we can't determine $\lambda$ exactly from the initial conditions; but we can determine it rather precisely by calculating the first few values of the sequences.

Here's Java code that does this. Here are the results up to $d=10$. (I'm not including $a_n$ and $s_n$ themselves, as these already have almost $200$ digits at $d=10$.)

\begin{array}{c|c|c|c|c|c|c} d&\frac{s_d}{a_d}&\frac{s_d-s_{d-1}}{a_d-a_{d-1}}&2^{-d}\left(\frac{s_d-s_{d-1}}{a_d-a_{d-1}}+1\right)\\\hline 1&0.5&1.0&1.0\\ 2&1.6&2.3333333333333333&0.8333333333333333\\ 3&4.038461538461538&4.619047619047619&0.7023809523809523\\ 4&9.063515509601181&9.2642089093702&0.6415130568356375\\ 5&19.126989287194817&19.141876050195236&0.6294336265686011\\ 6&39.25397857420277&39.254022488049124&0.6289691013757676\\ 7&79.50795714840554&79.50795714859716&0.6289684152234153\\ 8&160.01591429681108&160.01591429681108&0.6289684152219183\\ 9&321.0318285936221&321.0318285936221&0.6289684152219182\\ 10&643.0636571872442&643.0636571872442&0.6289684152219182\\ \end{array}

The calculation for $\lambda$ in the rightmost column converges very nicely. So the expected number of elements of a rooted ordered unlabeled binary tree of maximum depth exactly $d$ is approximately

$$ \lambda\cdot2^d-1 $$

with $\lambda\approx0.628968415221918$.

Thus your guess was already pretty good; you only underestimated the result by a factor of about $\frac54$.

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  • $\begingroup$ Thanks for the answer, very helpful. Slight correction, when u mention the constant $c$ it says $c^{2^d}$ instead of $c*2^d$ $\endgroup$ – Vk1 Jul 1 '18 at 7:19
  • $\begingroup$ @user13857: I'm not sure what you mean by "it says". Both the OEIS entry and my post say $c^{2^d}$, which is the correct form; I don't understand where or how you found $c\cdot2^d$. $\endgroup$ – joriki Jul 1 '18 at 7:29
  • $\begingroup$ My mistake, I confused it with something else :) But thanks for the clean and clear answer :D $\endgroup$ – Vk1 Jul 1 '18 at 8:28

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