Python:使用列表创建二叉搜索树

Question

我的代码的目标是从txt文件中获取每个单独的单词并将其放入一个列表中，然后使用该列表创建二叉搜索树来计算每个单词的频率并按字母顺序打印每个单词以及它的频率。中的每个单词只能包含字母、数字、- 或 ' 我无法用我的初学者编程知识来做的部分是使用我拥有的列表来制作二叉搜索树(我只能插入整个列表放在一个节点中，而不是将每个单词放在一个节点中来构建树)。到目前为止我的代码是这样的:

def read_words(filename):
    openfile = open(filename, "r")
    templist = []
    letterslist = []
    for lines in openfile:
        for i in lines:
            ii = i.lower()
            letterslist.append(ii)
    for p in letterslist:
        if p not in ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z',"'","-",' '] and p.isdigit() == False:
            letterslist.remove(p)      
    wordslist = list("".join(letterslist).split())
    return wordslist

class BinaryTree:
    class _Node:
        def __init__(self, value, left=None, right=None):
            self._left = left
            self._right = right
            self._value = value
            self._count = 1


    def __init__(self):
        self.root = None

    def isEmpty(self):
        return self.root == None

    def insert(self, value) :
        if self.isEmpty() :
            self.root = self._Node(value)
            return
        parent = None
        pointer = self.root
        while (pointer != None) :
            if value == pointer._value:
                pointer._count += 1
                return

            elif value < pointer._value:
                parent = pointer
                pointer = pointer._left

            else :
                parent = pointer
                pointer = pointer._right            

        if (value <= parent._value) :
            parent._left = self._Node(value)
        else :
            parent._right = self._Node(value)    

    def printTree(self):
        pointer = self.root
        if pointer._left is not None:
            pointer._left.printTree()
        print(str(pointer._value) + " " + str(pointer._count))
        if pointer._right is not None:
            pointer._right.printTree()




    def createTree(self,words):
        if len(words) > 0:
            for word in words:
                BinaryTree().insert(word)
            return BinaryTree()
        else:
            return None

    def search(self,tree, word):
        node = tree
        depth = 0
        count = 0
        while True:
            print(node.value)
            depth += 1
            if node.value == word:
                count = node.count
                break
            elif word < node.value:
                node = node.left
            elif word > node.value:
                node = node.right
        return depth, count


def main():
    words = read_words('sample.txt')
    b = BinaryTree()
    b.insert(words)
    b.createTree(words)
    b.printTree()

Answer 1

由于您是初学者，我建议使用递归而不是迭代来实现树方法，因为这会导致更简单的实现。虽然递归乍一看似乎是一个有点困难的概念，但通常它是最简单的方法。

这是二叉树的草稿实现，它使用递归来插入、搜索和打印树，它应该支持您需要的功能。

class Node(object):
    def __init__(self, value):
        self.value = value
        self.left = None
        self.right = None
        self.count = 1

    def __str__(self):
        return 'value: {0}, count: {1}'.format(self.value, self.count)

def insert(root, value):
    if not root:
        return Node(value)
    elif root.value == value:
        root.count += 1
    elif value < root.value:
        root.left = insert(root.left, value)
    else:
        root.right = insert(root.right, value)

    return root

def create(seq):
    root = None
    for word in seq:
        root = insert(root, word)

    return root

def search(root, word, depth=1):
    if not root:
        return 0, 0
    elif root.value == word:
        return depth, root.count
    elif word < root.value:
        return search(root.left, word, depth + 1)
    else:
        return search(root.right, word, depth + 1)

def print_tree(root):
    if root:
        print_tree(root.left)
        print root
        print_tree(root.right)

src = ['foo', 'bar', 'foobar', 'bar', 'barfoo']
tree = create(src)
print_tree(tree)

for word in src:
    print 'search {0}, result: {1}'.format(word, search(tree, word))

# Output
# value: bar, count: 2
# value: barfoo, count: 1
# value: foo, count: 1
# value: foobar, count: 1
# search foo, result: (1, 1)
# search bar, result: (2, 2)
# search foobar, result: (2, 1)
# search bar, result: (2, 2)
# search barfoo, result: (3, 1)