List of paths to nested dictionary
A simple recursion will do the work

def build_nested_helper(path, text, container):
    segs = path.split('/')
    head = segs[0]
    tail = segs[1:]
    if not tail:
        container[head] = lambda: msg(text)
    else:
        if head not in container:
            container[head] = {}
        build_nested_helper('/'.join(tail), text, container[head])

def build_nested(paths):
    container = {}
    for path in paths:
        build_nested_helper(path, path, container)
    return container

menu = build_nested(items)


squashed code: 159 chars

def f(p,t,c):
 s=p.split('/');a=s[0];b=s[1:]
 if b:
  if a not in c:c[a]={}
  f('/'.join(b),t,c[a])
 else:c[a]=lambda:msg(t)
menu={}
for i in items:f(i,i,menu)