Write a function is_bst, which takes a Tree t and returns True if, and only if, t is a valid binary search tree, which means that: Each node has at most two children (a leaf is automatically a valid binary search tree). The children are valid binary search trees. For every node, the entries in that node's left child are less than or equal to the label of the node. For every node, the entries in that node's right child are greater than the label of the node.其中class Tree: def init(self, label, branches=[]): for b in branches: assert isinstance(b, Tree) self.label = label self.branches = list(branches) def is_leaf(self): return not self.branches def map(self, fn): self.label = fn(self.label) for b in self.branches: b.map(fn) def contains(self, e): if self.label == e: return True for b in self.branches: if e in b: return True return False def repr(self): if self.branches: branch_str = ', ' + repr(self.branches) else: branch_str = '' return 'Tree({0}{1})'.format(self.label, branch_str) def str(self): def print_tree(t, indent=0): tree_str = ' ' * indent + str(t.label) + "\n" for b in t.branches: tree_str += print_tree(b, indent + 1) return tree_str return print_tree(self).rstrip()

下面是根据题目要求实现的is_bst函数代码：

def is_bst(t):
    def helper(t, low, high):
        if t is None:
            return True
        if not low <= t.label <= high:
            return False
        left_is_bst = helper(t.branches[0], low, t.label)
        right_is_bst = helper(t.branches[1], t.label, high)
        return left_is_bst and right_is_bst

    return helper(t, float('-inf'), float('inf'))

该函数中，helper函数是一个递归函数，用于判断二叉搜索树的合法性。对于每个节点t，它的左子树的所有节点都应该小于等于t的值，右子树的所有节点都应该大于t的值。在判断左子树和右子树时，需要传入当前节点t的值作为边界值。如果当前节点t为空，则直接返回True。最后，在is_bst函数中调用helper函数，并传入根节点t以及最小值和最大值的边界值float('-inf')和float('inf')，得到最终结果并返回。

需要注意的是，该函数实现中假设二叉搜索树中不存在重复元素。如果存在重复元素，则需要对判断条件进行修改。

相关问题

Write a function is_bst, which takes a Tree t and returns True if, and only if, t is a valid binary search tree, which means that: Each node has at most two children (a leaf is automatically a valid binary search tree). The children are valid binary search trees. For every node, the entries in that node's left child are less than or equal to the label of the node. For every node, the entries in that node's right child are greater than the label of the node.其中class Tree: def __init__(self, label, branches=[]): for b in branches: assert isinstance(b, Tree) self.label = label self.branches = list(branches) def is_leaf(self): return not self.branches def map(self, fn): self.label = fn(self.label) for b in self.branches: b.map(fn) def __contains__(self, e): if self.label == e: return True for b in self.branches: if e in b: return True return False def __repr__(self): if self.branches: branch_str = ', ' + repr(self.branches) else: branch_str = '' return 'Tree({0}{1})'.format(self.label, branch_str) def __str__(self): def print_tree(t, indent=0): tree_str = ' ' * indent + str(t.label) + "\n" for b in t.branches: tree_str += print_tree(b, indent + 1) return tree_str return print_tree(self).rstrip()

下面是 is_bst 函数的实现：

def is_bst(t):
    def is_bst_helper(t, min, max):
        if t is None:
            return True
        if t.label < min or t.label > max:
            return False
        left_bst = is_bst_helper(t.branches[0], min, t.label - 1)
        right_bst = is_bst_helper(t.branches[1], t.label + 1, max)
        return left_bst and right_bst

    return is_bst_helper(t, float('-inf'), float('inf'))

is_bst 函数使用递归来判断树 t 是否为二叉搜索树。is_bst_helper 函数是一个辅助函数，它接受三个参数：t 表示当前要判断的节点，min 表示当前节点值的下限，max 表示当前节点值的上限。如果当前节点为空，则返回 True。如果当前节点的值小于下限或大于上限，则返回 False。否则，递归检查左子树和右子树是否也是二叉搜索树，且左子树的值都小于等于当前节点的值，右子树的值都大于等于当前节点的值。最终，is_bst 函数调用 is_bst_helper 函数，并将根节点的下限和上限分别设置为负无穷和正无穷，来检查整棵树是否为二叉搜索树。

## Problem 7: Is BST Write a function `is_bst`, which takes a Tree `t` and returns `True` if, and only if, `t` is a valid binary search tree, which means that: - Each node has at most two children (a leaf is automatically a valid binary search tree). - The children are valid binary search trees. - For every node, the entries in that node's left child are less than or equal to the label of the node. - For every node, the entries in that node's right child are greater than the label of the node. An example of a BST is:  Note that, if a node has only one child, that child could be considered either the left or right child. You should take this into consideration. Hint: It may be helpful to write helper functions `bst_min` and `bst_max` that return the minimum and maximum, respectively, of a Tree if it is a valid binary search tree. ```python def is_bst(t): """Returns True if the Tree t has the structure of a valid BST. >>> t1 = Tree(6, [Tree(2, [Tree(1), Tree(4)]), Tree(7, [Tree(7), Tree(8)])]) >>> is_bst(t1) True >>> t2 = Tree(8, [Tree(2, [Tree(9), Tree(1)]), Tree(3, [Tree(6)]), Tree(5)]) >>> is_bst(t2) False >>> t3 = Tree(6, [Tree(2, [Tree(4), Tree(1)]), Tree(7, [Tree(7), Tree(8)])]) >>> is_bst(t3) False >>> t4 = Tree(1, [Tree(2, [Tree(3, [Tree(4)])])]) >>> is_bst(t4) True >>> t5 = Tree(1, [Tree(0, [Tree(-1, [Tree(-2)])])]) >>> is_bst(t5) True >>> t6 = Tree(1, [Tree(4, [Tree(2, [Tree(3)])])]) >>> is_bst(t6) True >>> t7 = Tree(2, [Tree(1, [Tree(5)]), Tree(4)]) >>> is_bst(t7) False """ "*** YOUR CODE HERE ***" ``` 。

根据你的问题，`is_bst` 函数存在问题，因为它只有一个参数。为了实现该函数，需要编写两个辅助函数 `bst_min` 和 `bst_max`，它们返回一个树的最小值和最大值。然后，可以使用这些帮助器函数来实现主函数 `is_bst`。以下是代码的修改版本：

def bst_min(t):
    """Returns the minimum value in a binary search tree."""
    if t.is_leaf():
        return t.label
    return min(t.label, bst_min(t.branches[0]), bst_min(t.branches[-1]))

def bst_max(t):
    """Returns the maximum value in a binary search tree."""
    if t.is_leaf():
        return t.label
    return max(t.label, bst_max(t.branches[-1]), bst_max(t.branches[0]))

def is_bst(t):
    """Returns True if the Tree t has the structure of a valid BST.
    >>> t1 = Tree(6, [Tree(2, [Tree(1), Tree(4)]), Tree(7, [Tree(7), Tree(8)])])
    >>> is_bst(t1)
    True
    >>> t2 = Tree(8, [Tree(2, [Tree(9), Tree(1)]), Tree(3, [Tree(6)]), Tree(5)])
    >>> is_bst(t2)
    False
    >>> t3 = Tree(6, [Tree(2, [Tree(4), Tree(1)]), Tree(7, [Tree(7), Tree(8)])])
    >>> is_bst(t3)
    False
    >>> t4 = Tree(1, [Tree(2, [Tree(3, [Tree(4)])])])
    >>> is_bst(t4)
    True
    >>> t5 = Tree(1, [Tree(0, [Tree(-1, [Tree(-2)])])])
    >>> is_bst(t5)
    True
    >>> t6 = Tree(1, [Tree(4, [Tree(2, [Tree(3)])])])
    >>> is_bst(t6)
    True
    >>> t7 = Tree(2, [Tree(1, [Tree(5)]), Tree(4)])
    >>> is_bst(t7)
    False
    """
    if t.is_leaf():
        return True
    elif len(t.branches) == 1:
        return is_bst(t.branches[0])
    else:
        left = t.branches[0]
        right = t.branches[-1]
        if bst_max(left) <= t.label < bst_min(right):
            return is_bst(left) and is_bst(right)
        else:
            return False