Cumulative Mul：编写一个函数cumulative_mul，它会改变Tree t，使每个节点的标签变成其标签的乘积

Question

编写一个函数cumulative_mul，它会改变Tree t，使每个节点的标签成为其标签与以该节点为根的子树中所有标签的乘积。 提示：仔细考虑何时对树进行变异以及该变异是否应该在处理子树之前或之后发生 类树： def init(自身，标签，分支=[]): 对于分支中的 b： 断言 isinstance(b, 树) self.label = 标签 self.branches = 列表（分支）

def is_leaf(self):
    return not self.branches

类树： def init(自身，标签，分支=[]): 对于分支中的 b： 断言 isinstance(b, 树) self.label = 标签 self.branches = 列表（分支）

def is_leaf(self):
    return not self.branches

def cumulative_mul(t):
    """Mutates t so that each node's label becomes the product of all labels in
    the corresponding subtree rooted at t.

    >>> t = Tree(1, [Tree(3, [Tree(5)]), Tree(7)])
    >>> cumulative_mul(t)
    >>> t
    Tree(105, [Tree(15, [Tree(5)]), Tree(7)])
    >>> otherTree = Tree(2, [Tree(1, [Tree(3), Tree(4), Tree(5)]), Tree(6, [Tree(7)])])
    >>> cumulative_mul(otherTree)
    >>> otherTree
    Tree(5040, [Tree(60, [Tree(3), Tree(4), Tree(5)]), Tree(42, [Tree(7)])])
    """
    "*** YOUR CODE HERE ***"
    if t.is_leaf():
        return
    for b in t.branches:
        cumulative_mul(b)
        if isinstance(b, Tree):
            t.label *= b.label

我的英语可能无法清楚地表达我的想法，请原谅我。 我可以大致了解结构，使用递归的基本思想是当它的叶子返回值时，当它是子树时返回子树标签的乘积。结构层层发生。但让我困惑的是，当您需要将子树的两个标签相乘时，它是如何工作的。

Answer 1

要处理来自不同

labels

的

subtrees

的乘法，您可以先计算每个

product

的

labels

的

subtree

，然后将其与当前的

node's label

相乘。

# class to represent the object for problem
class Tree: 
    def __init__(self, label, branches=[]): # init function to create object
        self.label = label
        self.branches = branches

    def is_leaf(self): # check the 'self' is leaf
        return not self.branches

# function to manipulate with object 'Tree'
def cumulative_mul(t):
    if t.is_leaf(): # check by using method 
        return

    subtree_product = 1  # Initialize product for the current subtree
    for b in t.branches: # cycle 
        cumulative_mul(b)
        if isinstance(b, Tree): # check
            subtree_product *= b.label  # Multiply labels of subtrees

    t.label *= subtree_product  # Multiply the current node's label by the subtree product

# Test cases for problem
t = Tree(1, [Tree(3, [Tree(5)]), Tree(7)])
cumulative_mul(t) # run test cases
print(t.label)  # Output: 105

otherTree = Tree(2, [Tree(1, [Tree(3), Tree(4), Tree(5)]), Tree(6, [Tree(7)])])
cumulative_mul(otherTree) # run test cases 
print(otherTree.label)  # Output: 5040