What is the average size of a jump in a binary tree? Assume that a full binary tree is layed out in memory recursively in the following way.
First the Root followed by Tree representation of left subtree followed by tree representation of right subtree.  Assume tree is complete and full.

If a search path requires to go from array location $a[i]$ to $a[k]$, then its referrred to as a jump and its of size $k-i$, since you travelled $k-i$ units of memory.

Given this arrangement of the tree, what is the expected size of a jump in this tree?
Consider a binary search tree with layers $0,\ldots,h$, where layer $\ell$ contains $2^\ell$ nodes. The left child of a node $x$ is located next to it (at distance 1). If $x$ is at layer $\ell$, then its left subtree contain $2^{h-\ell}-1$ nodes, and so the right child of $x$ is at distance $2^{h-\ell}$. These observations are illustrated in the figure:



Here $h = 3$, and the root ($\ell = 0$) has a left child at distance $1$ and a right child at distance $2^{h-\ell} = 8$.

We will calculate the average size of a jump in a random root-to-leaf path. This corresponds to searching for a random leaf.

The expected jump at level $\ell$ is thus $\frac{1+2^{h-\ell}}{2}$. Since there are $h$ jumps overall (at levels $0$ to $h-1$), the average jump is
$$
\frac{1}{h} \sum_{\ell=0}^{h-1} \frac{1+2^{h-\ell}}{2} =
\frac{1}{2} + \frac{1 + \cdots + 2^{h-1}}{h} =
\frac{1}{2} + \frac{2^h-1}{h}.
$$
Since the number of nodes in the tree is $n := 2^{h+1}-1$, the average jump is $\Theta(n/\log n)$. The same holds for the number of leaves $m := 2^h$.