I want to render this data structure as an unordered list.
menu = [
[1, 0],
[2, 1],
[3, 1],
[4, 3],
[5, 3],
[6, 5],
[7,1]
]
[n][0] is the key
[n][1] references the parent keyThe desired output is:
<ul>
<li>Node 1</li>
<ul>
<li>Node 2</li>
<li>Node 3</li>
<ul>
<li>Node 4</li>
<li>Node 5</li>
<ul>
<li>Node 6</li>
</ul>
</ul>
<li>Node 7</li>
</ul>
</ul>
I could probably do this without recursion but that would be no fun. What is the most efficient way to solve this problem 开发者_高级运维with recursion?
Thanks!
def render(nodes, parent = 0):
if parent not in nodes:
return
print('<ul>')
for n in nodes[parent]:
print('<li>Node %d</li>' % n)
render(nodes, n)
print('</ul>')
Here is the output
>>> nodes = {}
>>> for n in menu:
if n[1] not in nodes:
nodes[n[1]] = []
nodes[n[1]].append(n[0])
>>> render(nodes)
<ul>
<li>Node 1</li>
<ul>
<li>Node 2</li>
<li>Node 3</li>
<ul>
<li>Node 4</li>
<li>Node 5</li>
<ul>
<li>Node 6</li>
</ul>
</ul>
<li>Node 7</li>
</ul>
</ul>
I would not use two-element lists, even if your structure is that simple. Use some TreeNode
class, and give it an appropriate __str__
method, e.g.
class TreeNode(object):
# ...
# methods for adding children (instances of TreeNode again) etc.
def __str__(self):
ret = "<li>%s" % self.value
if self.children:
children = "".join([str(c) for c in self.children])
ret += "<ul>%s</ul>" % children
ret += "</li>"
return ret
... or something like that. Didn't test it, though. Enclose the whole tree's representation in an <ul>
tag.
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