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我正在尝试编写一个创建二进制搜索树的函数。这是我到目前为止的内容：

def add_items(bst, low, high):
    if low == high:
        bst.insert(high)
        return
    else:
        left = add_items(bst, low, high)
        right = add_items(bst, low, high)
        item = BinarySearchTreeMap.Item(low)
        node = BinarySearchTreeMap.BinarySearchTreeMap.Node(item)
        node.left = left
        node.right = right
        return node

我注意到的问题是该函数在完成所有递归调用时返回一个节点。我想返回此二进制搜索树的根，但是我不确定如何最后返回它。图片包括对我正在尝试做的更详细的描述。我很感谢任何人提供的任何帮助，建议或想法。实际上，add_items有点像一个辅助函数，它被以下代码片段调用：

def create_complete_bst(n):
    bst = BinarySearchTreeMap.BinarySearchTreeMap()
    add_items(bst, 1, n)
    return bst

P.S。这是我已经提供并正在该程序中使用的二进制搜索树类

class BinarySearchTreeMap:

class Item:
    def __init__(self, key, value=None):
        self.key = key
        self.value = value


class Node:
    def __init__(self, item):
        self.item = item
        self.parent = None
        self.left = None
        self.right = None

    def num_children(self):
        count = 0
        if (self.left is not None):
            count += 1
        if (self.right is not None):
            count += 1
        return count

    def disconnect(self):
        self.item = None
        self.parent = None
        self.left = None
        self.right = None


def __init__(self):
    self.root = None
    self.size = 0

def __len__(self):
    return self.size

def is_empty(self):
    return len(self) == 0


# raises exception if not found
def __getitem__(self, key):
    node = self.find(key)
    if (node is None):
        raise KeyError(str(key) + " not found")
    else:
        return node.item.value

# returns None if not found
def find(self, key):
    curr = self.root
    while (curr is not None):
        if (curr.item.key == key):
            return curr
        elif (curr.item.key > key):
            curr = curr.left
        else:  # (curr.item.key < key)
            curr = curr.right
    return None


# updates value if key already exists
def __setitem__(self, key, value):
    node = self.find(key)
    if (node is None):
        self.insert(key, value)
    else:
        node.item.value = value

# assumes key not in tree
def insert(self, key, value=None):
    item = BinarySearchTreeMap.Item(key, value)
    new_node = BinarySearchTreeMap.Node(item)
    if (self.is_empty()):
        self.root = new_node
        self.size = 1
    else:
        parent = self.root
        if(key < self.root.item.key):
            curr = self.root.left
        else:
            curr = self.root.right
        while (curr is not None):
            parent = curr
            if (key < curr.item.key):
                curr = curr.left
            else:
                curr = curr.right
        if (key < parent.item.key):
            parent.left = new_node
        else:
            parent.right = new_node
        new_node.parent = parent
        self.size += 1


# raises exception if key not in tree
def __delitem__(self, key):
    node = self.find(key)
    if (node is None):
        raise KeyError(str(key) + " is not found")
    else:
        self.delete_node(node)

# assumes key is in tree + returns value assosiated
def delete_node(self, node_to_delete):
    item = node_to_delete.item
    num_children = node_to_delete.num_children()

    if (node_to_delete is self.root):
        if (num_children == 0):
            self.root = None
            node_to_delete.disconnect()
            self.size -= 1

        elif (num_children == 1):
            if (self.root.left is not None):
                self.root = self.root.left
            else:
                self.root = self.root.right
            self.root.parent = None
            node_to_delete.disconnect()
            self.size -= 1

        else: #num_children == 2
            max_of_left = self.subtree_max(node_to_delete.left)
            node_to_delete.item = max_of_left.item
            self.delete_node(max_of_left)

    else:
        if (num_children == 0):
            parent = node_to_delete.parent
            if (node_to_delete is parent.left):
                parent.left = None
            else:
                parent.right = None

            node_to_delete.disconnect()
            self.size -= 1

        elif (num_children == 1):
            parent = node_to_delete.parent
            if(node_to_delete.left is not None):
                child = node_to_delete.left
            else:
                child = node_to_delete.right

            child.parent = parent
            if (node_to_delete is parent.left):
                parent.left = child
            else:
                parent.right = child

            node_to_delete.disconnect()
            self.size -= 1

        else: #num_children == 2
            max_of_left = self.subtree_max(node_to_delete.left)
            node_to_delete.item = max_of_left.item
            self.delete_node(max_of_left)

    return item

# assumes non empty subtree
def subtree_max(self, curr_root):
    node = curr_root
    while (node.right is not None):
        node = node.right
    return node


def inorder(self):
    for node in self.subtree_inorder(self.root):
        yield node

def subtree_inorder(self, curr_root):
    if(curr_root is None):
        pass
    else:
        yield from self.subtree_inorder(curr_root.left)
        yield curr_root
        yield from self.subtree_inorder(curr_root.right)

def __iter__(self):
    for node in self.inorder():
        yield node.item.key

Answer 1

经过修改后的建议：

def create_complete_bst(n):
    bst = BinarySearchTreeMap.BinarySearchTreeMap()
    add_items(bst, 1, n)
    return bst


def add_items(bst, low, high):
    if low == high:
        bst.insert(high)
        return
    elif high < low:
        return
    else:
        mid = (low+high) // 2
        bst.insert(mid)
        add_items(bst, low, mid-1)
        add_items(bst, mid+1, high)
        return bst

递归创建二进制搜索树的功能

1 个答案: