Find nearest link with BeautifullSoup (python)

I am doing a small project where I extract occurences of political leaders in newspapers. Sometimes a politician will be mentioned, and there is neither a parent or child with a link. (due I guess to semantically bad markup).

So I want to create a function that can find the nearest link, and then extract that. In the case below the search string is Rasmussen and the link I want is: /307046.

#-*- coding: utf-8 -*-

from bs4 import BeautifulSoup
import re

tekst = '''
<li>
  <div class="views-field-field-webrubrik-value">
    <h3>
      <a href="/307046">Claus Hjort spiller med mrkede kort</a>
    </h3>
  </div>
  <div class="views-field-field-skribent-uid">
    <div class="byline">Af: <span class="authors">Dennis Kristensen</span></div>
  </div>
  <div class="views-field-field-webteaser-value">
    <div class="webteaser">Claus Hjort Frederiksens argumenter for at afvise
      trepartsforhandlinger har ikke hold i virkeligheden. Hans rinde er nok
      snarere at forberede det ideologiske grundlag for en Løkke Rasmussens
      genkomst som statsministe
    </div>
  </div>
  <span class="views-field-view-node">
    <span class="actions">
      <a href="/307046">Ls mere</a>
      |
      <a href="/307046/#comments">Kommentarer (4)</a>
    </span>
  </span>
</li>
'''

to_find = "Rasmussen"
soup = BeautifulSoup(tekst)
contexts = soup.find_all(text=re.compile(to_find)) 

def find_nearest(element, url, direction="both"):
    """Find the nearest link, relative to a text string.
    When complete it will search up and down (parent, child),
    and only X levels up down. These features are not implemented yet.
    Will then return the link the fewest steps away from the
    original element. Assumes we have already found an element"""

    # Is the nearest link readily available?
    # If so - this works and extracts the link.
    if element.find_parents('a'):
        for artikel_link in element.find_parents('a'):
            link = artikel_link.get('href')
            # sometimes the link is a relative link - sometimes it is not
            if ("http" or "www") not in link:
                link = url+link
                return link
    # But if the link is not readily available, we will go up
    # This is (I think) where it goes wrong
    # ↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
    if not element.find_parents('a'):
        element =  element.parent
        # Print for debugging
        print element #on the 2nd run (i.e <li> this finds <a href=/307056> 
        # So shouldn't it be caught as readily available above?
        print u"Found: %s" % element.name
        # the recursive call
        find_nearest(element,url)

# run it
if contexts:
    for a in contexts:
        find_nearest( element=a, url="http://information.dk")

The direct call below works:

print contexts[0].parent.parent.parent.a['href'].encode('utf-8')

For reference the whole sorry code is on bitbucket: https://bitbucket.org/achristoffersen/politikere-i-medierne

(p.s. Using BeautifullSoup 4)

EDIT: SimonSapin asks me to define nearest: By nearest I mean the link that are the fewest nesting levels away from the search term, in either direction. In the text above, the a href produced by the drupal-based newspaper site, is neither a direct parent or child of the tag where the search string is found. So BeautifullSoup Can't find it.

I suspect a 'fewest charachters' away would often work too. In that case a soulution could be hacked together with find and rfind - but I would really like to do this via BS. Since this would work: contexts[0].parent.parent.parent.a['href'].encode('utf-8') it must be possible to generalise that to a script.

EDIT: Maybe I should emphasize that I'am looking for a BeautifulSoup solution. Combining BS with a custom/simpel breath-first-search as suggested by @erik85 would quickly become messy I think.