[英]Python - Converting an n-ary tree to a binary tree
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。 如果树中的一个节点有两个以上的子节点,则必须将后代最多的子节点分配为左节点,后代数目第二大的子节点分配为右子节点。 其余子项的添加如下:1)接收具有最大后代数的子项2)将其添加到“左/右”节点。 那个时候孩子少的那个。
*如果在任何时候我都需要选择后代数量最多的子代,但后代数量相同的有两个+,则我选择键值较大的子代。
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
DSM的评论帮助我了解了发生的情况。 在convert_to_binary_tree
方法的第一部分中选择left_child
和right_child
之后,您没有将它们从子级列表中删除。 这意味着以后,当您将当前节点的所有子代添加到新的父代中时,您将向左或向右添加自己的子代(或彼此添加)。 当您递归到那些孩子时,您可能会无限循环。
我不太了解您选择left_child
和right_child
的逻辑,因此我没有固定的代码可以建议您。 一个快速但丑陋的解决方法是将if child in (left_child, right_child): continue
for循环顶部的if child in (left_child, right_child): continue
语句,在该循环中您将其他子项分配给新的父项。
请注意,当前代码中还有另一个错误,其中左右子级的后代计数将变得不正确。 那是因为当您将一些以前的兄弟姐妹作为孩子推入他们时,您并没有更新计数。
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