Question

我正在尝试定义3个函数，这些函数将进入一个函数，然后将返回True或False，具体取决于给定列表（树）中是否存在特定的数字（键）。这是我得到的代码：

def is_empty_tree(tree):
    return isinstance(tree, list) and not tree

def is_leaf(tree):
    return isinstance(tree, int)

def left_subtree(tree):
    return tree[0]

def right_subtree(tree):
    return tree[2]

def get_key(tree):
    return tree[1]

def is_innernode(tree):
    return isinstance(tree,list)

def func_traverse(binary_tree, func1, func2, func3):
    if is_empty_tree(binary_tree):
        return func3()
    if is_leaf(binary_tree):
        return func2(binary_tree)
    if is_innernode(binary_tree):
        return func3(get_key(binary_tree), \
         traverse(left_subtree(binary_tree), inner_node_fn, leaf_fn, \
         empty_tree_fn), traverse(right_subtree(binary_tree), \ 
         inner_node_fn, leaf_fn, empty_tree_fn))

def func3():
    return False

def func2(num):
    return num

def func3(binary_tree, left, right):
    if isinstance(binary_tree, int):
        return exists(binary_tree)
    else:
         ????

def contain__the_key(key, binary_tree):
    return key == func_traverse(binary_tree, search, exists, not_in_tree)

#Test example
contain_the_key(4, [6, 2, [[4, 3, 9], 0, []]])
>>> True
conatin_the_key(6, [[], 1, 5])
>>> False

问题是我对如何编写func1，func2，func3并没有真正的好主意，因此如果它在列表中，它将返回true，否则返回false。任何想法将不胜感激！

检查密钥是否在带有嵌套列表的树中

0 个答案: