Question

在这种情况下，它要求我进行间接递归。用于“ def count_less”功能。这意味着计算小于给定值的节点数。对于我的代码，我将具有attributeError。谁能帮我检查我的代码？

class BST:
    """A Binary Search Tree."""

    def __init__(self, container=[]):
        """(BST, list) -> NoneType
        Initialize this BST by inserting the items from container (default [])
        one by one, in the order given.
        """
        # Initialize empty tree.
        self.root = None
        # Insert every item from container.
        for item in container:
            self.insert(item)

    def __str__(self):
        """(BST) -> str
        Return a "sideways" representation of the values in this BST, with
        right subtrees above nodes above left subtrees and each value preceded
        by a number of TAB characters equal to its depth.
        """
        if self.root:
            return self.root._str("")
        else:
            return ""
    def count_less(self, item):
        """(BST, object) -> int
        Return the number of items in this BST that are strictly less than
        item.
        """
        if self.root:
            return self.root.count_less(item)
        else:
            return 0

我需要在“ _BSTNode”类中编写函数体，并在“ BST”类中调用该类方法。

class _BSTNode:
    """A node in a BST."""

    def __init__(self, item, left=None, right=None):
        """(_BSTNode, object, _BSTNode, _BSTNode) -> NoneType
        Initialize this node to store item and have children left and right.
        """
        self.item = item
        self.left = left
        self.right = right

    def _str(self, indent):
        """(_BSTNode, str) -> str
        Return a "sideways" representation of the values in the BST rooted at
        this node, with right subtrees above nodes above left subtrees and each
        value preceded by a number of TAB characters equal to its depth, plus
        indent.
        """
        if self.right:
            right_str = self.right._str(indent + "\t")
        else:
            right_str = ""
        if self.left:
            left_str = self.left._str(indent + "\t")
        else:
            left_str = ""
        return right_str + indent + str(self.item) + "\n" + left_str

    def count_less(self: '_BSTNode', item: object) -> int:
        """
        Return the number of items in the BST rooted at this node that are
        strictly less than item.
        """
        if not self.item:
            return 0
        elif item <= self.item:
            if self.left:
                return 1 + self.left.count_less(item)
            return 0
        elif self.item < item:
            if self.left and self.right:
                return 1 + self.left.count_less(item) + self.right.count_less(item)
            elif self.left and not self.right:
                return 1 + self.left.count_less(item)
            elif self.right and not self.left:
                return 1 + self.right.count_less(item)
            else:
                return 1

这是我输入的示例，输出中出现错误。

    >>> t = BST(container=[5,1,2,3,4,6,7,8,9])
    >>> t.count_less(10)

Answer 1

您可以依靠None的虚假性质，并使用if语句检查子节点是否存在。我非常确定以下代码不会返回正确的答案：如果self.item == item会发生什么？左侧和右侧子级的子级都有可能小于item参数。

elif item < self.item:
    if self.left:
        return 1 + self.left.count_less(item)
    return 1
elif item > self.item:
    if self.right:
        return 1 + self.right.count_less(item)
    return 1

BST中节点数的间接递归小于给定值

1 个答案: