  Given two binary trees on $N$ labeled leaves, the quartet distance between
the trees is the number of disagreeing quartets. By permuting the leaves at
random, the expected quartets distance between the two trees is
$\frac{2}{3}\binom{N}{4}$. However, no strongly explicit construction reaching
this bound asymptotically was known.
  We consider complete, balanced binary trees on $N=2^n$ leaves, labeled by $n$
long bit sequences. Ordering the leaves in one tree by the prefix order, and in
the other tree by the suffix order, we show that the resulting quartet distance
is $\left(\frac{2}{3} + o(1)\right)\binom{N}{4}$, and it always exceeds the
$\frac{2}{3}\binom{N}{4}$ bound.
