How to traverse a binary Tree with a recursive generator?

Question

I am trying to traverse a Binary Tree which is created in the following code. to be precise, the Binary Tree is a class and should include an iterator calling another function namely inorder(). this method should be a recursive generator and yield the value of nodes in every iteration.I tried to create a dictionary to follow the nodes but when I try to call the inorder() method, it doesn't work. Is there any missing point that I don't know? I used while and it creates the dictionary of left side of tree (it is a clumsy way). please help me accomplish this code.

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()

I changed my code again I accomplished the code but I'm sure it is not the best way to traverse a Binary tree. I defined a method -traverse()- in my class and returned a list of nodes in order now (which at first wasn't ordered, so I used sort() method.) then I made a loop over this list in my generator, inorder() function, to yield the element of it. All your comments are very welcome to optimize the code. please recommend a proper solution based on the specific Tree class in this code.

Solution 1

Perhaps I'm missing something, but I'm not sure why the dictionary is relevant in inorder(). Think about what an in-order traversal looks like in general:

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

and so in terms of generators, this would look like:

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

Solution 2

I am thoroughly confused by your thinking. For one thing, there's not actually any dictionaries in this code, and I don't understand why you introduced the d global.

All you need to do for in-order traversal of a binary tree is to traverse the left, the current label, and the right:

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

That's it.

However I would make some improvements to your code:

# 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()