class Node: #定义节点类 def add_child(self, child): assert isinstance(child, Node) is_leaf = False if child in self.children: return self.children.append(child) hashes = [] for node in self.children: hashes.append(node.get_hash()) prehash = ''.join(hashes) self.node_hash = md5(prehash.encode('utf-8')).hexdigest()

这段代码定义了一个Node类，表示区块链中的节点。其中包含一个add_child方法，用于向当前节点添加子节点。添加子节点时，首先判断子节点是否已经存在于当前节点的子节点列表中，如果存在则不进行操作，否则将子节点添加到子节点列表中。然后计算子节点列表中所有节点的哈希值，并将它们拼接起来得到一个新的哈希值。最后使用MD5算法对新的哈希值进行加密，得到一个长度为32的哈希字符串，并将其赋值给当前节点的node_hash属性。这样就完成了节点的添加和哈希计算操作。需要注意的是，为了确保子节点类型正确，add_child方法中使用assert语句进行了类型检查，只有当子节点是Node类型时才能进行添加操作。

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。

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()

Here's one possible implementation of the `is_bst` function using recursion to check the validity of each node in the binary search tree:

def is_bst(t):
    def is_bst_helper(t, min_val, max_val):
        if t.is_leaf():
            return True
        elif len(t.branches) == 1:
            return False
        else:
            left_bst = is_bst_helper(t.branches[0], min_val, t.label)
            right_bst = is_bst_helper(t.branches[1], t.label, max_val)
            return left_bst and right_bst and min_val <= t.label and t.label <= max_val
    
    return is_bst_helper(t, float('-inf'), float('inf'))

The `is_bst_helper` function takes in a tree `t`, a minimum value `min_val`, and a maximum value `max_val`, and returns `True` if `t` is a valid binary search tree within the range `[min_val, max_val]`. If `t` is a leaf, it is automatically a valid binary search tree, so `True` is returned. If `t` has only one child, it cannot be a valid binary search tree, so `False` is returned. Otherwise, the function recursively checks that the left and right subtrees are valid binary search trees within the appropriate ranges, and that the label of `t` is between `min_val` and `max_val`. Finally, the `is_bst` function calls `is_bst_helper` with the root of the input tree `t` and the initial range `[-inf, inf]`.