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