Question

我正在编写一个算法来计算树所拥有的节点总数。这是代码：

def tree_nodes(tree):
    """Given a tree, returns the total amount of nodes it has."""
    def recursive_node_count(node):
        """Recursive count function."""
        childs = tree[node]
        if childs == None:
            return 1
        else:
            nodes_so_far = 0
            for i in xrange(len(childs)):
                nodes_in_this_branch = recursive_node_count(childs[i])
                nodes_so_far += nodes_in_this_branch          
            return nodes_so_far

    root = tree['root']
    total_nodes = recursive_node_count(root)
    return total_nodes

tree基本上是列表字典。例如：

tree = {0: [1], 1: [2], 2: [3, 4], 3: None, 4: None, 'root': 0}

enter image description here

当我尝试运行我的代码时，这是我收到的输出：

at Answer.py. not in a function on line 31
at Answer.py. in recursive_node_count on line 31
at Answer.py. in recursive_node_count on line 31
at Answer.py. in recursive_node_count on line 31
at Answer.py. in recursive_node_count on line 31
at Answer.py. in recursive_node_count on line 31
at Answer.py. in recursive_node_count on line 31
at Answer.py. in recursive_node_count on line 31
at Answer.py. in recursive_node_count on line 31
at Answer.py. in recursive_node_count on line 31
at Answer.py. in tree_nodes on line 36
at Answer.py. in <module> on line 96

这些是原始代码中的行31（在函数定义内），36（在定义内）和96（对定义的调用）：

31: nodes_in_this_branch = recursive_node_count(childs[i])
36: total_nodes = recursive_node_count(root)
96: nodes = tree_nodes(tree)

我检查了语法，缩进，制表符，空格，但我找不到错误。我是Python的初学者。你能帮助我吗？

Answer 1

您当前的代码存在2个问题，

您目前不计算根节点
当没有孩子时你返回1，在这种情况下你应该返回0。我们同样需要计算每个级别的孩子数量，因此应该根据孩子的长度来初始化nodes_so_far。列表

纠正这些，该功能变为：

def tree_nodes(tree):
    """Given a tree, returns the total amount of nodes it has."""
    def recursive_node_count(node):
        """Recursive count function."""
        childs = tree[node]
        if childs == None:
            return 0 # There are no child so return 0 in base case
        else:
            nodes_so_far = len(childs) # set to number of nodes passed
            for i in xrange(len(childs)):
                nodes_in_this_branch = recursive_node_count(childs[i])
                nodes_so_far += nodes_in_this_branch          
            return nodes_so_far
    root = tree['root']
    total_nodes = 1 + recursive_node_count(root) # Add 1 to count the root node
    return total_nodes

在干运行中，这给出了输出：

>>> tree = {0: [1], 1: [2], 2: [3, 4], 3: None, 4: None, 'root': 0}
>>> tree_nodes(tree)
5

Answer 2

你的答案遗漏了几件事。这是围绕代码编写的固定版本：

def tree_nodes(tree):
    def recursive_node_count(node):
        if node is None:
            return 0
        total_nodes = 1
        if tree[node]:
            for child in tree[node]:
                if child:
                    total_nodes += recursive_node_count(child)
        return total_nodes

    root = tree['root']
    total_nodes = recursive_node_count(root)
    return total_nodes

>>> tree = {0: [1], 1: [2], 2: [3, 4], 3: None, 4: None, 'root': 0}
>>> print tree_nodes(tree)
5

Answer 3

根据您提供的代码，我无法重现错误。语法上没有任何问题，确实运行良好。如果你能更深入地了解你所得到的错误，我会编辑我的答案。

话虽如此，如mu所述，此代码将返回错误数量的节点。具体来说，它将返回树的叶数，因为如果它没有任何子节点，您只计算当前节点。可以通过将nodes_so_far初始化为表示当前节点的1来解决此问题。

作为建议，您可能希望将for in xrange python语句切换为普通for in语句。 for in语句遍历列表，因此您不必使用索引号索引回列表。

下面的代码说明了这些变化。即使在只有一个节点且是根节点的情况下，此代码也将始终输出正确数量的节点。

def tree_nodes(tree):
    """Given a tree, returns the total amount of nodes it has."""
    def recursive_node_count(node):
        """Recursive count function."""
        childs = tree[node]
        if not childs:
            # Return 1 for the current node.
            return 1
        else:
            # Initialize to 1 to count the current node.
            nodes_so_far = 1
            # Python for in statement
            for child in childs:
                nodes_for_child = recursive_node_count(child)
                nodes_so_far += nodes_for_child
            return nodes_so_far
    root = tree['root']
    total_nodes = recursive_node_count(root)
    return total_nodes

print(tree_nodes(tree={0: [1], 1: [2], 2: [3, 4], 3: None, 4: None, 'root': 0}))
print(tree_nodes(tree={0: None, 'root': 0}))

递归函数中的“不在函数中”错误

3 个答案: