Python二叉搜索树与双向链表转换实现方法

# encoding=utf8
'''
题目：输入一棵二叉搜索树，将该二叉搜索树转换成一个排序的双向链表。
要求不能创建任何新的结点，只能调整树中结点指针的指向。
'''
class BinaryTreeNode():
  def __init__(self, value, left = None, right = None):
    self.value = value
    self.left = left
    self.right = right
def create_a_tree():
  node_4 = BinaryTreeNode(4)
  node_8 = BinaryTreeNode(8)
  node_6 = BinaryTreeNode(6, node_4, node_8)
  node_12 = BinaryTreeNode(12)
  node_16 = BinaryTreeNode(16)
  node_14 = BinaryTreeNode(14, node_12, node_16)
  node_10 = BinaryTreeNode(10, node_6, node_14)
  return node_10
def print_a_tree(root):
  if root is None:return
  print_a_tree(root.left)
  print root.value, ' ',
  print_a_tree(root.right)
def print_a_linked_list(head):
  print 'linked_list:'
  while head is not None:
    print head.value, ' ',
    head = head.right
  print ''
def create_linked_list(root):
  '''构造树的双向链表，返回这个双向链表的最左结点和最右结点的指针'''
  if root is None:
    return (None, None)
  # 递归构造出左子树的双向链表
  (l_1, r_1) = create_linked_list(root.left)
  left_most = l_1 if l_1 is not None else root
  (l_2, r_2) = create_linked_list(root.right)
  right_most = r_2 if r_2 is not None else root
  # 将整理好的左右子树和root连接起来
  root.left = r_1
  if r_1 is not None:r_1.right = root
  root.right = l_2
  if l_2 is not None:l_2.left = root
  # 由于是双向链表，返回给上层最左边的结点和最右边的结点指针
  return (left_most, right_most)
if __name__ == '__main__':
  tree_1 = create_a_tree()
  print_a_tree(tree_1)
  (left_most, right_most) = create_linked_list(tree_1)
  print_a_linked_list(left_most)
  pass