This is the more important case of "even" lattices, where all inner products are integral and all vector norms are even. We follow pages 131-134 in Wolfgang Ebeling, Lattices and Codes, available for sale at [LINK][1] 
\\$45.00 for paperback.

Given an even positive lattice $\Lambda$ with covering radius below $\sqrt 2,$ form the integral Lorentzian lattice $L =  \Lambda \oplus U.$ Elements are of the form 
$ (\lambda, m,n)   $ where $\lambda \in \Lambda, \; m,n \in \mathbb Z.$ The norm on $L$ is given by
$$   (\lambda, m,n)^2 = \lambda^2 + 2 mn.$$ We infer the inner product 
$$ (\lambda_1, m_1,n_1) \cdot (\lambda_2, m_2,n_2) = \lambda_1  \cdot \lambda_2 + m_1 n_2 + m_2 n_1.$$
 

We choose a particular set of roots (elements of norm 2) beginning with any $\lambda \in \Lambda$ by
$$ \tilde{\lambda} =   \left( \lambda, 1, 1 - \frac{\lambda^2}{2} \right).$$ 

Let the group $ R \subseteq \mbox{Aut}(L)$ be generated by reflections in all the $\tilde{\lambda}$ and by $\pm 1.$

Given any $ l \in L,$ a primitive null vector, we have $l \cdot l = 0,$ and so $l \in l^\perp.$ Furthermore, if 
$k \in l^\perp$ as well, then $ (k + l)^2 = k^2.$ As a result, we may form the lattice spanned $\langle l \rangle$ by $l$ itself, then form another lattice with norm, $$ E(l) = l^\perp / \langle l \rangle.$$ It took me a bit of doing to confirm (a calculus exercise along with Cauchy-Schwarz) that $E(l)$ is positive definite, all norms are positive except for the $0$ class.

Given any root $r \in L,$ meaning $r^2 = 2,$ we get the reflection $$ s_r (z) = z - ( r \cdot z) r.$$
Also $ s_r^2(z) = z.$

It is an exercise to show that, when $ s_r(z) = y,$ then $E(z)$ and $E(y)$ are isomorphic. 

There is rather more than first appears to the question:

http://mathoverflow.net/questions/70666/lorentzian-characterization-of-genus 

In particular, every even lattice in the same genus as  $\Lambda$ occurs as some  $E(u),$ where
$u  \in L$ is a primitive null vector. Furthermore, taking $w = (0, 0,1),$ we find that
$E(w) = \Lambda.$

So, all we really need to do to prove class number one is show that every primitive null vector can be taken to $w = (\vec{0}, 0,1),$ 
by a sequence of operations in $R.$

Given some primitive null vector
$$ z = (\xi, a,b),$$
so that $ 2 a b = - \xi^2.$ Note that, in order to have a primitive null vector, if one of $a,b$ is 0, then $\xi = 0$ and the other one of $a,b$ is $\pm 1.$

In the first case, suppose $|b| < |a|.$ Then $ | 2 a b| = \xi^2 < 2 a^2,$ so in fact
$$ \left( \frac{\xi}{a} \right)^2 < 2.$$
We choose the root $ \tilde{\lambda} =   \left( 0, 1, 1 \right).$ Then $z \cdot \tilde{\lambda} = b + a,$ and
$$  s_{\tilde{\lambda}} (z) = z - (  \tilde{\lambda} \cdot z)  \tilde{\lambda} =   (\xi, -b,-a).$$ 


Therefore, we may always force the second case, which is $|a| \leq |b|.$ We assume that $a \neq 0,$ so that
$ b = \frac{- \xi^2}{2a}.$ Now we have
$ 2 a^2 \leq  | 2 a b| = \xi^2,$ so $ \left( \frac{\xi}{a} \right)^2 \geq 2.$ From the covering radius condition, there is then some 
nonzero vector $\lambda \in \Lambda$ such that the rational number
$$   \left( \frac{\xi}{a} - \lambda \right)^2 < 2.   $$

We form the root from this  $\lambda,$ as in
$$ \tilde{\lambda} =   \left( \lambda, 1, 1 - \frac{\lambda^2}{2} \right).$$
For convenience we write
$$  a' = \frac{a}{2} \left( \frac{\xi}{a} - \lambda \right)^2  $$
From the equation $z \cdot \tilde{\lambda} = a - a',$ we see that $a' \in \mathbb Z.$


Well, we have   $$  s_{\tilde{\lambda}} (z) =  (\xi - ( a - a') \lambda, a',b'), $$ where
$$ b' = b - ( a - a') \left( 1 - \frac{\lambda^2}{2} \right) = \frac{- \xi^2}{2a} - ( a - a') \left( 1 - \frac{\lambda^2}{2} \right).$$    


Now, if $a'=0,$ then $ s_{\tilde{\lambda}} (z) = (0,0,\pm 1),$ and an application of $\pm 1 \in R$ takes us to
$w = (0, 0,1),$ with
$E(w) = \Lambda.$

If, instead, $a' \neq 0,$ note that our use of the covering radius condition shows that
$ | a'| < | a|,$ while $a,a'$ share the same $\pm$ sign, that is their product is positive. So $a - a'$ shares the same sign, and
$| a - a'| < |a|.$  From $2 a b = - \xi^2$ we know that $b$ has the opposite sign. But $2 a' b' = - (\xi - ( a - a') \lambda)^2,$ so
$b'$ has the opposite sign to $a'$ and $a-a'$ and the same sign as $b.$ Now, $\lambda^2 \geq 2,$ so
$$ 1 - \frac{\lambda^2}{2} \leq 0,   $$ 
and $  ( a - a') \left( 1 - \frac{\lambda^2}{2} \right)$ has the same sign as $b$ and $b'.$
 From
$$ b' = b - ( a - a') \left( 1 - \frac{\lambda^2}{2} \right)$$ 
we conclude that $| b'| \leq |b|.$

So, these steps have $|a| + |b|$ strictly decreasing, until such time that one of them becomes 0, and we have arrived at $w.$ So, actually, the covering radius hypothesis implies that all primitive null vectors are mapped to $w$ by a finite sequence of reflections, so that all the $E(z)$ are in fact isomorphic to $E(w) = \Lambda.$ That is, all lattices in the genus are in fact isomorphic, and the class number is one. 


  [1]: http://www.ams.org/bookstore-getitem/item=VWALM-9