Python-将n元树转换为二叉树

Question

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()

I have to convert a given tree into a binary tree. If a node in the tree has more than 2 children, I have to assign the child with the most descendants as the left node and the child with the second largest number of descendents as the right child. The remaining children are added as following: 1) Take child with largest number of descendants 2) Add it to Left/Right node. Whichever has fewer children at that time.

*If at any time I need to select the child with the largest number of descendants, but there are two+ with the same number of descendents, I take the one with the larger key value.

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 

Answer 1

DSM's comment helped me see what is going on. After you pick left_child and right_child in the first parts of your convert_to_binary_tree method, you're not removing them from the list of children. This means that later, when you go to add all of the current node's children into new parents, you're adding the left and right children to themselves (or each other). When you recurse into those children, you can end up infinitely looping.

I don't really understand the logic of your left_child and right_child selections, so I don't have fixed code to suggest to you. A quick but ugly fix would be to put a if child in (left_child, right_child): continue statement at the top of the for loop where you're assigning the other children to new parents.

Note that there's another bug in your current code, where the descendent counts for the left and right children will become incorrect. That's because you're not updating the count when you push some of their former siblings into them as children.