You can start from the leaf and climb up:
select p_n || ' - ' || n
from t
where p_n is not null
start with n = 11
connect by prior p_n = n
order by level desc
EDIT: OK, this makes things a little bit more complicated...
You can climb up from both nodes but then you'll have to remove the duplicates paths (for example between 6 and 3 there is no need to go via the root 1)
try something like this:
select the_path
from
(select case when connect_by_root(n) = 11 then p_n || ' - ' || n else n || ' - ' || p_n end the_path,
count(*) over (partition by n, p_n) cnt
from t
where p_n is not null
start with n in (11, 4)
connect by prior p_n = n)
where cnt = 1;