# encoding=utf8
'''
題目：輸入一棵二叉搜索樹，將該二叉搜索樹轉(zhuǎn)換成一個排序的雙向鏈表。
要求不能創(chuàng)建任何新的結(jié)點，只能調(diào)整樹中結(jié)點指針的指向。
'''
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):
  '''構(gòu)造樹的雙向鏈表，返回這個雙向鏈表的最左結(jié)點和最右結(jié)點的指針'''
  if root is None:
    return (None, None)
  # 遞歸構(gòu)造出左子樹的雙向鏈表
  (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
  # 將整理好的左右子樹和root連接起來
  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
  # 由于是雙向鏈表，返回給上層最左邊的結(jié)點和最右邊的結(jié)點指針
  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