3

我正在尝试遍历在以下代码中创建的二叉树。准确地说，二叉树是一个类，应该包括一个调用另一个函数的迭代器，即 inorder()。这个方法应该是一个递归生成器，并在每次迭代中产生节点的值。我试图创建一个字典来跟踪节点，但是当我尝试调用 inorder() 方法时，它不起作用。有什么我不知道的遗漏点吗？我使用了while，它创建了树左侧的字典（这是一种笨拙的方式）。请帮我完成这段代码。

d=[]

# A binary tree class.
class Tree(object):
    def __init__(self, label, left=None, right=None):
        self.label = label
        self.left = left
        self.right = right
        self.d=dict()
    def __repr__(self, level=0, indent="    "):
        s = level * indent + self.label
        if self.left:
            s = s + "\n" + self.left.__repr__(level + 1, indent)
        if self.right:
            s = s + "\n" + self.right.__repr__(level + 1, indent)
        return s

def traverse(self):
    if self.left:
        lastLabel=self.label
        self.left.traverse()
    if self.right:
        lastLabel=self.label
        d.append(lastLabel)
        self.right.traverse()
    else:
        d.append(self.label)
    return d

def __iter__(self):
    return inorder(self)

# Create a Tree from a list.
def tree(sequence):
    n = len(sequence)
    if n == 0:
        return []
    i = n / 2
    return Tree(sequence[i], tree(sequence[:i]), tree(sequence[i+1:]))

# A recursive generator that generates Tree labels in in-order.
def inorder(t):
    for i in range(len(d)):
        yield d[i]    

def test(sequence):
# Create a tree.
    t = tree(sequence)
# Print the nodes of the tree in in-order.
    result = []
    for x in t:
        result.append(x)
    print x
    print

    result_str = ''.join(result)

# Check result
    assert result_str == sequence
    del d[:]
def main():
    # Third test
    test("0123456789")

    print 'Success! All tests passed!'

if __name__ == '__main__':
    main()

我再次更改了代码 我完成了代码，但我确信这不是遍历二叉树的最佳方法。我在我的类中定义了一个方法 -traverse()- 并现在按顺序返回了一个节点列表（起初没有排序，所以我使用了 sort() 方法。）然后我在我的生成器中对该列表进行了循环, inorder() 函数，以产生它的元素。非常欢迎您的所有意见来优化代码。请根据此代码中的特定 Tree 类推荐适当的解决方案。

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2 回答 2

8

也许我遗漏了一些东西，但我不确定为什么字典与inorder(). 想想一般的有序遍历是什么样子的：

def inorder(t):
    # Process left sub tree
    # Process t
    # Process right sub tree

所以就生成器而言，这看起来像：

def inorder(t):
    if t.left:
        for elem in inorder(t.left):
            yield elem
    yield t
    if t.right:
        for elem in inorder(t.right):
            yield elem
于 2013-02-11T16:01:58.617 回答
2

我完全被你的想法弄糊涂了。一方面，这段代码中实际上没有任何字典，我不明白你为什么要引入d全局。

对于二叉树的中序遍历，您需要做的就是遍历左侧、当前标签和右侧：

def inorder(tree):
    for label in tree.left:
        yield label
    yield tree.label
    for label in tree.right:
        yield label

就是这样。

但是，我会对您的代码进行一些改进：

# Document classes and functions with docstrings instead of comments
class Tree(object):
    """A binary tree class"""
    def __init__(self, label, left=None, right=None):
        """Label is the node value, left and right are Tree objects or None"""
        self.label = label
        self.left = left   # Tree() or None
        self.right = right # Tree() or None

    def __repr__(self):
        return 'Tree(%r, %r, %r)' % (self.label, self.left, self.right)

    def __iter__(self):
        # No need for a separate inorder() function
        if self.left is not None:
            for t in self.left:
                yield t
        yield self.label
        if self.right is not None:
            for t in self.right:
                yield t

def tree(indexable):
    """Return a tree of anything sliceable"""
    ilen = len(indexable)
    if not ilen:
        # You should be clearer about empty values
        # left and right should be Tree (something with left, right, and __iter__)
        # or None if there is no edge.
        return None 
    center = ilen // 2 # floor division
    return Tree(indexable[center], tree(indexable[:center]), tree(indexable[center+1:]))


def test():
    seq = range(10)
    t = tree(seq)
    # list(t) will consume an iterable
    # no need for "result = []; for x in t: result.append(x)"
    assert seq == list(t)


if __name__ == '__main__':
    test()
于 2013-02-12T20:46:30.010 回答