慕桂英546537
您在这段代码中提到了两个问题： tree = Tree(k, d[k]) #1st issue: should input a Tree() as 2nd parameter
for child in d[k]:
build_tree(child) #2nd issue: should populate tree, not iterate recursively over children keys您可以通过本质上将for循环移至第二个参数来解决它们，以列表理解的形式并泼溅该列表，使它们成为参数。然后确保您的递归函数返回创建的树： return Tree(k,
*[build_tree(child) for child in d[k]]
)更多想法与您的问题无关，但这里还有一些您可以使用的想法。建议您的代码成为一个可以作为A参数传递给的函数，这样字典的作用域就只是该函数的本地作用域，而不会影响全局作用域。由于此功能与类密切相关Tree，因此最好将其定义为类中的静态方法或类方法。当您拥有树的（子、父）元组时，这些元组隐式定义哪个节点是根，因此您可以省略将文字 66 传递给函数。该函数应该能够自行找出哪个是根。在创建字典时，它还可以收集哪些节点有父节点。根就是不在该集合中的节点。所以把所有这些放在一起你会得到这个：from collections import defaultdictclass Tree: def __init__(self, node, *children): self.node = node self.children = children if children else [] def __str__(self): return "%s" % (self.node) def __repr__(self): return "%s" % (self.node) def __getitem__(self, k): if isinstance(k, int) or isinstance(k, slice): return self.children[k] if isinstance(k, str): for child in self.children: if child.node == k: return child def __iter__(self): return self.children.__iter__() def __len__(self): return len(self.children) @classmethod def from_pairs(Cls, pairs): # Turn pairs into nested dictionary d = defaultdict(list) children = set() for child, parent in pairs: d[parent].append(child) # collect nodes that have a parent children.add(child) # Find root: it does not have a parent root = next(parent for parent in d if parent not in children) # Build nested Tree instances recursively from the dictionary def subtree(k): return Cls(k, *[subtree(child) for child in d[k]]) return subtree(root)# Sample runA = [(61, 66), (50, 61), (68, 61), (33, 61), (57, 66), (72, 66), (37, 68), (71, 33), (6, 50), (11, 37), (5, 37)]tree = Tree.from_pairs(A)
开满天机
你很接近了。关键是将新节点返回给父节点并将其追加到父节点的子节点列表中。如果您的父级列表在初始化时是固定的，只需使用临时列表，然后在访问并创建子级后创建父级。这是一个最小的例子：from collections import defaultdict, namedtupledef build_tree(tree, root): if root: return Node(root, [build_tree(tree, x) for x in tree.get(root, [])])def print_tree(root, indent=0): if root: print(" " * indent + str(root.val)) for child in root.children: print_tree(child, indent + 2)if __name__ == "__main__": A = [(61, 66), (50, 61), (68, 61), (33, 61), (57, 66), (72, 66), (37, 68), (71, 33), (6, 50), (11, 37), (5, 37)] Node = namedtuple("Node", "val children") nodes = defaultdict(list) for child, parent in A: nodes[parent].append(child) print_tree(build_tree(nodes, 66))输出：66 61 50 6 68 37 11 5 33 71 57 72
万千封印
这是了解可重用模块和相互递归的机会。首先，我们将定义特定于(id, parent)输入结构形状的函数 -# main.pydef id(node): return node[0]def parent(node): return node[1]n = (12,34)id(n) # => 12parent(n) # => 34也许你知道根节点是66，但这对我们的程序来说很难推断，但对我们来说很容易定义。让我们明确地包含(66, None)在您的输入数据中，其中parent=None表示根节点-A = \ [ (61, 66), (50, 61), (68, 61), (33, 61) , (57, 66), (72, 66), (37, 68), (71, 33) , (6, 50), (11, 37), (5, 37), (66, None) # don't forget root node, 66 ]现在我们可以使用该tree模块轻松构建我们的树 -# main.pyfrom tree import treedef id #...def parent #...A = [ ... ]B = tree \ ( A # list of nodes , parent # foreign key , lambda node, children: # node reconstructor (id(node), children(id(node))) # primary key )print(B)# [(66, [(61, [(50, [(6, [])]), (68, [(37, [(11, []), (5, [])])]), (33, [(71, [])])]), (57, []), (72, [])])]请注意，如何tree不关心输入的形状；可以使用任何节点结构。该tree功能很灵活，允许我们构造与输入节点完全不同的形状的树节点 -# main.pyfrom tree import treefrom json import dumpsdef id #...def parent #...A = [ ... ]C = tree \ ( A , parent , lambda node, children: dict([("id", id(node)), ("children", children(id(node)))]) )print(dumps(C))[ { "id": 66 , "children": [ { "id": 61 , "children": [ { "id": 50 , "children": [ { "id": 6, "children": [] } ] } , { "id": 68 , "children": [ { "id": 37 , "children": [ { "id": 11, "children": [] } , { "id": 5, "children": [] } ] } ] } , { "id": 33 , "children": [ { "id": 71, "children": [] } ] } ] } , { "id": 57, "children": [] } , { "id": 72, "children": [] } ] }]现在我们可以看看 的实现tree。请注意如何tree不对输入节点的形状做出任何假设 -# tree.pyfrom index import index, getdef empty(): return []def tree (all, indexer, maker, root = None): mem = index(all, indexer) def many(all): return list(map(one, all)) def one(single): return maker(single, lambda r: many(get(mem, r, empty()))) return many(get(mem, root))我们的实现tree依赖于另一个模块，index. 将数据结构（例如索引）以及对这些数据结构进行操作的函数进行分组是在模块之间划定界限的好方法。这里也没有对输入形状做出任何假设 -# index.pyfrom functools import reducedef empty(): return {}def update(t, k, f): if k in t: return { **t, k: f(get(t, k)) } else: return { **t, k: f() }def get(t, k, default = None): if k in t: return t[k] else: return defaultdef append(t, k, v): return update(t, k, lambda r = []: [ *r, v ])def index(ls, indexer): return reduce \ ( lambda t, v: append(t, indexer(v), v) , ls , empty() )通过在浏览器中运行来验证我们的结果：run this program on repl.it