# Show that the average depth of a leaf in a binary tree with n vertices is $\Omega(\log n)$.

Let $T$ be a tree with$n$ vertices, having height $h$. If there are any internal vertices in $T$ at levels less than $h — 1$ that do not have two children, take a leaf at level $h$ and move it to be such a missing child. This only lowers the average depth of a leaf in this tree, and since we are trying to prove a lower bound on the average depth, it suffices to prove the bound for the resulting tree. Repeat this process until there are no more internal vertices of this type. As a result, all the leaves are now at levels $h — 1$ and $h$. Now delete all vertices at level $h$. This changes the number of vertices by at most (one more than) a factor of two and so has no effect on a big-Omega estimate (it changes $\log n$ by at most 1). Now the tree is complete, and it has $2^{h-1}$ leaves, all at depth $h — 1$, where now $n = 2^{h-1}$. The desired estimate follows. The statement above is the answer from the textbook,but I couldn't understand it. Is there anyone can give me a more explicit answer?

• Can you be more precise about which parts you understand and which parts you don't? May 18, 2015 at 13:11
• **As a result, all the leaves are now at levels h—1 and h ...... Now the tree is complete, ..... ** @Kimball May 19, 2015 at 2:03

We can do much better than this lower bound and compute the exact asymptotics.

Remark. I just realized I computed the average depth of the internal nodes and not the leaves. Correction TBA.

Note that the functional equation for binary trees classified by height is $$T(z, u) = 1 + uzT(uz, u)^2.$$

As we are interested in the average we need to compute $$G(z) = \left.\frac{\partial}{\partial u} T(z, u)\right|_{u=1}.$$

We differentiate the functional equation, getting $$\frac{\partial}{\partial u} T(z, u) = z T(uz, u)^2 + 2uz T(z, u) \left(z\frac{\partial}{\partial z} T(z, u) + \frac{\partial}{\partial u} T(z, u)\right).$$

Writing $T(z)$ for the solution of $T(z) = 1 + z T(z)^2$, the generating function of the Catalan numbers $$C_n = \frac{1}{n+1} {2n\choose n}$$ we set $u=1$ to obtain $$G(z) = zT(z)^2 + 2z T(z) (z T'(z) + G(z))$$ which yields $$G(z) = z\frac{T(z)^2+2z T(z) T'(z)}{1-2z T(z)}.$$

From the functional equation we have $$T'(z) = T(z)^2 + 2z T(z) T'(z)$$ or $$T'(z) = \frac{T(z)^2}{1-2zT(z)}$$ so $G(z)$ becomes $$G(z) = z\frac{T(z)^2+2z T(z)^3/(1-2zT(z)) }{1-2z T(z)} \\ = z\frac{T(z)^2-2zT(z)^3+2z T(z)^3}{(1-2z T(z))^2} \\ = z\frac{T(z)^2}{(1-2z T(z))^2}.$$

To extract coefficients from this we use Lagrange inversion in the integral $$[z^n] G(z) = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} z\frac{T(z)^2}{(1-2z T(z))^2} \; dz.$$

Put $w=T(z)$ so that $w = 1 + z w^2$ or $$z = \frac{w-1}{w^2} \quad\text{and}\quad dz = \left(\frac{1}{w^2}-2\frac{w-1}{w^3}\right) \; dw.$$

This gives for the integral $$\frac{1}{2\pi i} \int_{|w-1|=\epsilon} \frac{w^{2n}}{(w-1)^n} \frac{w^2}{(1-2(w-1)/w)^2} \left(\frac{1}{w^2}-2\frac{w-1}{w^3}\right) \; dw \\ = \frac{1}{2\pi i} \int_{|w-1|=\epsilon} \frac{w^{2n}}{(w-1)^n} \frac{w^4}{(w-2(w-1))^2} \left(\frac{1}{w^2}-2\frac{w-1}{w^3}\right) \; dw \\ = \frac{1}{2\pi i} \int_{|w-1|=\epsilon} \frac{w^{2n}}{(w-1)^n} \frac{w^4}{(1-(w-1))^2} \frac{1-(w-1)}{w^3} \; dw \\ = \frac{1}{2\pi i} \int_{|w-1|=\epsilon} \frac{w^{2n+1}}{(w-1)^n} \frac{1}{1-(w-1)}\; dw.$$

Expanding into a series we have $$\frac{1}{2\pi i} \int_{|w-1|=\epsilon} \frac{1}{(w-1)^n} \sum_{q=0}^{2n+1} {2n+1\choose q} (w-1)^q \sum_{p=0}^\infty (w-1)^p \; dw$$

which yields $$\sum_{q=0}^{n-1} {2n+1\choose q} = -{2n+1\choose n} + \sum_{q=0}^{n} {2n+1\choose q} = -{2n+1\choose n} + 2^{2n}.$$

This gives the sequence $$1, 6, 29, 130, 562, 2380, 9949, 41226, 169766,\ldots$$ which is OEIS A008549 where we learn of additional combinatorial interpretations.

For the average we get $$-\frac{1}{n} C_n^{-1} {2n+1\choose n} + \frac{1}{n} 2^{2n} C_n^{-1}.$$

The second term produces the dominant asymptotics while the first term yields $$-\frac{n+1}{n} \frac{(n!)^2}{(2n)!} \frac{(2n+1)!}{n!(n+1)!} = -\frac{2n+1}{n} = -2 - \frac{1}{n}.$$

Now using the asymptotics of the Catalan numbers which are $$C_n\sim \frac{4^n}{n^{3/2}\sqrt{\pi}}$$

we finally obtain for the average depth the value $$\frac{1}{n} 2^{2n} \frac{n^{3/2}\sqrt{\pi}}{4^n}$$ which simplifies to $$\sqrt{\pi n}.$$

Note that the functional equation for binary trees classified by total height of the leaves is (use the fact that a tree on $n$ internal nodes has $n+1$ leaves) $$T(z, u) = 1 + u^2 z T(uz, u)^2.$$

As we are interested in the average we need to compute $$G(z) = \left.\frac{\partial}{\partial u} T(z, u)\right|_{u=1}.$$

We differentiate the functional equation, getting $$\frac{\partial}{\partial u} T(z, u) = 2u z T(uz, u)^2 + 2u^2 z T(z, u) \left(z\frac{\partial}{\partial z} T(z, u) + \frac{\partial}{\partial u} T(z, u)\right).$$

We set $u=1$ to obtain $$G(z) = 2z T(z)^2 + 2z T(z) (zT'(z)+G(z))$$ which yields $$G(z) = z\frac{2T(z)^2 + 2zT(z)T'(z)}{1-2zT(z)}.$$

Substituting in the formula for $T'(z)$ yields $$G(z) = z\frac{2T(z)^2 + 2zT(z)^3/(1-2zT(z))}{1-2zT(z)} \\ = z\frac{2T(z)^2 -4zT(z)^3 + 2zT(z)^3}{(1-2zT(z))^2} \\ = z\frac{2T(z)^2 - 2zT(z)^3}{(1-2zT(z))^2} \\ = 2z T(z)^2\frac{1 - zT(z)}{(1-2zT(z))^2}.$$

This time the Lagrange inversion integral is $$[z^n] G(z) = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} 2z T(z)^2\frac{1 - zT(z)}{(1-2zT(z))^2} \; dz.$$

Use the same substitution as before to obtain for the integral $$\frac{1}{2\pi i} \int_{|w-1|=\epsilon} \frac{2w^{2n}}{(w-1)^n} w^2 \frac{1-(w-1)/w}{(1-2(w-1)/w)^2} \left(\frac{1}{w^2}-2\frac{w-1}{w^3}\right) \; dw \\ = \frac{1}{2\pi i} \int_{|w-1|=\epsilon} \frac{2w^{2n}}{(w-1)^n} w \frac{1}{(1-2(w-1)/w)^2} \frac{1-(w-1)}{w^3} \; dw \\ = \frac{1}{2\pi i} \int_{|w-1|=\epsilon} \frac{2w^{2n}}{(w-1)^n} \frac{1}{(1-(w-1))^2} (1-(w-1)) \; dw \\ = \frac{1}{2\pi i} \int_{|w-1|=\epsilon} \frac{2w^{2n}}{(w-1)^n} \frac{1}{1-(w-1)} \; dw.$$

Re-write this to prepare for coefficient extraction: $$\frac{2}{2\pi i} \int_{|w-1|=\epsilon} \frac{1}{(w-1)^n} \sum_{q=0}^{2n} {2n\choose q} (w-1)^q \sum_{p=0}^\infty (w-1)^p \; dw.$$

This gives $$2\sum_{q=0}^{n-1} {2n\choose q} = 2^{2n} - {2n\choose n}.$$

This is the sequence $$0, 2, 10, 44, 186, 772, 3172, 12952, 52666, 213524,\ldots$$ which is OEIS A068551.

For the asymptotics of the lower order term we get $$-\frac{1}{n}{2n\choose n} C_n^{-1} = -\frac{n+1}{n} = -1 - \frac{1}{n}.$$

For the dominant term we get the same as before $$\frac{1}{n} 2^{2n} \frac{n^{3/2}\sqrt{\pi}}{4^n}$$ which simplifies to $$\sqrt{\pi n}.$$

Concrete verification. These data can be verified with the combstruct package (Maple). This is the code.

with(combstruct);

gf :=
proc(n)
option remember;
local trees, leaves;

trees := { T=Union(V, Prod(Z, Sequence(T, card=2))),
Z=Atom, V=Atom };

leaves :=
proc(struct, height)
if type(struct, function) then
if op(0, struct) = Sequence then
q=1..nops(struct));
else
q=1..nops(struct));
fi;
fi;

if struct = Z then return 0 fi;
return height;
end;

add(u^leaves(t, 0), t in allstructs([T, trees], size=2*n+1));
end;

f := n -> subs(u=1, diff(gf(n), u));



The output is

> seq(f(n), n=1..8);
2, 10, 44, 186, 772, 3172, 12952, 52666


which confirms the data from above.

A slightly different approach may be consulted at this MSE link.

What is written is not quite correct.

First off, you need to assume something like your tree is binary, because if it is a star graph, all vertices, except the root, have degree 1.

Now the idea of the argument is, given a binary tree, balance it out. This means move the leaves of the tree higher up until one gets a tree of some minimal height, call it $h$. (In the OP, $h$ is overused, and cannot mean both the initial height and the height after doing this process--e.g., think of a path graph (i.e., a tree with only one vertex at each depth).) This process will only lower the average depth of the tree, and as in the link, now $h$ is within one of $\log_2(n)$.