使用生成器进行二叉树有序遍历

Question

Description

I want my Binary tree to be iterable so that I can loop though it to visit every node once. Also, inorder is a generator function which returns the Iterator and hence satisfies the Iterable contract. But my code below instead of yielding every node just yields root node which in this case is A . What I am doing wrong?

Code

from collections import namedtuple

Node = namedtuple('Node', 'data, left, right')

root = Node('A',
        Node('B', Node('D', None, None), Node('E', None, None)),
        Node('C', None, Node('F', None, None)))

class BinaryTree(object):
    def __init__(self, root=None):
        self.root = root

    def __iter__(self):
        return self.inorder(self.root)

    def inorder(self, node):
        if node is None:
            return
        if node.left is not None:
            self.inorder(node.left)
        yield node
        if node.right is not None:
            self.inorder(node.right)

bt = BinaryTree(root)            
for node in bt:
    print node.data

Reference

https://stackoverflow.com/a/6916433/4260745

Answer 1

The problem is that you call self.inorder(node.left) , but don't do anything with the result, and as a result, the code is simply executed (well executed lazily, which means not executed), and the runtime environment continues to the next line.

You need to resolve it by propagating (ie re-yielding ) the elements that are generated by the calls to inorder :

    def inorder(self, node):
        if node is None:
            return
        if node.left is not None:
            for x in self.inorder(node.left) :
                # you need to *re-yield* the elements of the left and right child
                yield x
        yield node
        if node.right is not None:
            for x in self.inorder(node.right) :
                yield x