class Tree:
def __init__(self, new_key):
self.__key = new_key # Root key value
self.__children = [] # List of children
self.__num_of_descendants = 0 # Number of Descendants of this node
# Prints the given tree
def printTree(self):
return self.printTreeGivenPrefix("", True)
# Prints the given tree with the given prefix for the line
# last_child indicates whether the node is the last of its parent"s child
# or not
def printTreeGivenPrefix(self, line_prefix, last_child):
print(line_prefix, end="")
if last_child:
print("â””--> ", end="")
else:
print("|--> ", end="")
print(self.__key)
if len(self.__children) > 0:
next_pre = line_prefix
if last_child:
next_pre += " "
else:
next_pre += "| "
for child_index in range(len(self.__children)-1):
self.__children[child_index].\
printTreeGivenPrefix(next_pre, False)
self.__children[-1].printTreeGivenPrefix(next_pre, True)
def __repr__(self):
return "[" + str(self.__key) + "".join(
[ repr(child) for child in self.__children ]) + "]"
# This static function will load a tree with the format of below:
# [root[child_1][child_2]...[child_n]]
# Each child_i can be a tree with the above format, too
# pos is the position in the given string
@staticmethod
def loadTree(tree_str, pos = 0):
new_node = None
while pos < len(tree_str):
if tree_str[pos] == "[":
pos += 1
new_node = Tree(tree_str[pos])
while pos < len(tree_str) and tree_str[pos + 1] != "]":
pos += 1
child_tree, pos = Tree.loadTree(tree_str, pos)
if child_tree:
new_node.__children.append(child_tree)
new_node.__num_of_descendants += \
1 + child_tree.__num_of_descendants
return new_node, pos + 1
else:
pos += 1
return new_node, pos
def find_largest(self):
if self.__num_of_descendants == 1:
return self.__children[0]
else:
largest_child = self.__children[0]
for child in self.__children:
if child.__num_of_descendants > \
largest_child.__num_of_descendants:
largest_child = child
if child.__num_of_descendants == \
largest_child.__num_of_descendants:
if child.__key > largest_child.__key:
largest_child = child
return largest_child
def convert_to_binary_tree(self):
if self.__num_of_descendants != 0:
if self.__num_of_descendants < 3:
for child in self.__children:
child.convert_to_binary_tree()
if self.__num_of_descendants > 2:
left_child = self.__children[0]
for child in self.__children[1:]:
if len(child.__children) > len(left_child.__children):
left_child = child
elif len(child.__children) == len(left_child.__children):
if child.__key > left_child.__key:
left_child = child
self.__children.remove(left_child)
self.__num_of_descendants -= 1
right_child = self.__children[0]
for child in self.__children[1:]:
if len(child.__children) > len(right_child.__children):
right_child = child
elif len(child.__children) == len(right_child.__children):
if child.__key > right_child.__key:
right_child = child
self.__children.remove(right_child)
self.__num_of_descendants -= 1
print(self.__num_of_descendants)
print(self.__children)
print(left_child)
print(right_child)
#Move remaining children two either left_child or right_child.
while self.__num_of_descendants != 0:
largest_child = self.find_largest()
print(largest_child)
if left_child.__num_of_descendants < \
right_child.__num_of_descendants:
left_child.__children.append(largest_child)
left_child.__num_of_descendants += 1
self.__children.remove(largest_child)
self.__num_of_descendants -= 1
elif left_child.__num_of_descendants > \
right_child.__num_of_descendants:
right_child.__children.append(largest_child)
right_child.__num_of_descendants += 1
self.__children.remove(largest_child)
self.__num_of_descendants -= 1
elif left_child.__num_of_descendants == \
right_child.__num_of_descendants:
if left_child.__key > right_child.__key:
left_child.__children.append(largest_child)
left_child.__num_of_descendants += 1
self.__children.remove(largest_child)
self.__num_of_descendants -= 1
else:
right_child.__children.append(largest_child)
right_child.__num_of_descendants += 1
self.__children.remove(largest_child)
self.__num_of_descendants -= 1
#Now run recursion on left and right binary children.
self.__children.append(left_child)
self.__children.append(right_child)
self.__num_of_descendants = 2
print(self.__children)
for child in self.__children:
child.convert_to_binary_tree()
def main():
tree, processed_chars = Tree.loadTree('[z[y][x][w][v]]]')
tree.convert_to_binary_tree()
tree.printTree()
print(tree)
if __name__ == "__main__":
main()
我必须将给定的树转换为二叉树。如果树中的节点有两个以上的子节点，我必须将具有最多后代的子节点分配为左节点，将具有第二大子节点的子节点分配为右子节点。其余的孩子添加如下： 1）带走后代人数最多的孩子 2）将其添加到左/右节点。当时的孩子较少。
*如果我在任何时候需要选择后代数量最多的孩子，但是有两个+具有相同数量的后代，我会选择具有较大键值的孩子。
I get a print out like this...
2 #Number of 'z' children after left and right node chosen.
[[w], [v]] #Children of 'z'
[y] #Binary left child of 'z'
[x] #Binary right child of 'z'
[w] #This is a bug. It should be choosing 'v' as larger child of 'z' and assigning it to left child 'y'
[v] #This is a bug. see above.
[[y[w]], [x[v]]] #These are the children of node 'z'
â””--> z #schematic of binary tree
|--> y
| â””--> w
â””--> x
â””--> v
[z[y[w]][x[v]]] #final binary tree
答案 0 :(得分：0)
left_child方法的第一部分中选择right_child和convert_to_binary_tree后，您并未将其从子列表中删除。这意味着稍后，当您将所有当前节点的孩子添加到新父母中时，您需要将左右孩子添加到自己（或彼此）。当你进入那些孩子时，你可以最终无限循环。
我并不完全理解您left_child和right_child选项的逻辑，所以我没有固定的代码可以向您推荐。一个快速但丑陋的解决方法是在for循环的顶部放置一个if child in (left_child, right_child): continue语句，您可以将其他子项分配给新父母。
请注意，您当前的代码中还有另一个错误，即左右儿童的后代计数将变得不正确。这是因为当你把他们的一些兄弟姐妹当作孩子推进时，你并没有更新计数。