我有n个列表A，B，C ......其中包含a，b，c ...元素。我正在使用它们来创建可能的组合列表（列表列表），这样第一个元素取自表A，第二个取自B等。将这个平面结构转换为遵循根的树的最佳方法是什么通过列表A的元素，每个元素都包含列表B等的所有元素，从而以根路径的形式创建每个可能的组合？
答案 0 :(得分：3)
因此，如果您有[1, 2, 3]，[4, 5, 6]，[7, 8, 9]列表，那么您有this list of possible permutiations
所以我现在将浏览列表及其元素。所有代表您想要的树的路径。因此，如果所有内容都放入树中，我将能够找到节点1，然后转到4，然后在第三级找到7。如果我不这样做，我必须在那里添加。
paths = [
[1, 4, 7], [1, 4, 8], [1, 4, 9], [1, 5, 7], [1, 5, 8], [1, 5, 9], [1, 6, 7], [1, 6, 8], [1, 6, 9], [2, 4, 7], [2, 4, 8], [2, 4, 9], [2, 5, 7], [2, 5, 8], [2, 5, 9], [2, 6, 7], [2, 6, 8], [2, 6, 9], [3, 4, 7], [3, 4, 8], [3, 4, 9], [3, 5, 7], [3, 5, 8], [3, 5, 9], [3, 6, 7], [3, 6, 8], [3, 6, 9]
]
class Tree(object):
def __init__(self, node=None):
self.node = node
self.children = []
def addChild(self, child):
self.children.append(child)
def toList(self):
if len(self.children) > 0:
return [self.node, [x.toList() for x in self.children]]
else:
return [self.node]
tree = Tree()
# iterate through the lists
for path in paths:
subtree = tree
for value in path:
# check whether the current value already exists at this position in the tree
found = False
for child in subtree.children:
if value == child.node:
subtree = child
found = True
break
# attach the leaf to the current position in the tree if needed
if not found:
newchild = Tree(node=value)
subtree.addChild(newchild)
subtree = newchild
# use the found or created leaf as a position for the next value in the path-list
print tree.toList()
如果树看起来是对称的，你也可以只切片树的层次，给你列表A，B和C.然后，你回到原点，可以建立树：< / p>
paths = [
[1, 4, 7], [1, 4, 8], [1, 4, 9], [1, 5, 7], [1, 5, 8], [1, 5, 9], [1, 6, 7], [1, 6, 8], [1, 6, 9], [2, 4, 7], [2, 4, 8], [2, 4, 9], [2, 5, 7], [2, 5, 8], [2, 5, 9], [2, 6, 7], [2, 6, 8], [2, 6, 9], [3, 4, 7], [3, 4, 8], [3, 4, 9], [3, 5, 7], [3, 5, 8], [3, 5, 9], [3, 6, 7], [3, 6, 8], [3, 6, 9]
]
lists = [[]]*len(paths[0])
for i in xrange(len(paths[0])):
lists[i] = set([])
for l in paths:
lists[i].add(l[i])
def iterate(listnumber):
result = []
for element in lists[listnumber]:
if listnumber < len(lists)-1:
result.append([element, iterate(listnumber+1)])
else:
result.append([element])
return result
tree = iterate(0)
print tree
它遍历一堆列表（a，b，c）中的每一层，并将唯一成员存储在列表中。然后它有列表A，B，C并从中生成树。
这是我得到的树，0是人工根。节点被“回收”，因为它们都具有相同的内容。 （用dot制作。）
