Python二叉搜索树与双向链表转换算法示例

class TreeNode:
  def __init__(self, x):
    self.val = x
    self.left = None
    self.right = None
class Solution:
  def lastElem(self, list):
    if len(list) == 0:
      return None
    else: return list[len(list) - 1]
  def ConvertCore(self, pRoot, doubleLinkedList):
    if pRoot:
      if pRoot.left:
        self.ConvertCore(pRoot.left, doubleLinkedList)
      pRoot.left = self.lastElem(doubleLinkedList)
      if self.lastElem(doubleLinkedList):
        self.lastElem(doubleLinkedList).right = pRoot
      doubleLinkedList.append(pRoot)
      if pRoot.right:
        self.ConvertCore(pRoot.right, doubleLinkedList)
  def Convert(self, pRootOfTree):
    if pRootOfTree == None:
      return None
    doubleLinkedList = []
    self.ConvertCore(pRootOfTree, doubleLinkedList)
    return doubleLinkedList[0]

class TreeNode:
  def __init__(self, x):
    self.val = x
    self.left = None
    self.right = None
class Solution:
  def ConvertCore(self, pRoot, lastListNode):
    if pRoot:
      if pRoot.left:
        self.ConvertCore(pRoot.left, lastListNode)
      pRoot.left = lastListNode[0]
      if lastListNode[0]:
        lastListNode[0].right = pRoot
      lastListNode[0] = pRoot
      if pRoot.right:
        self.ConvertCore(pRoot.right, lastListNode)
  def Convert(self, pRootOfTree):
    # write code here
    if pRootOfTree == None:
      return None
    lastListNode = [None]
    self.ConvertCore(pRootOfTree, lastListNode)
    while lastListNode[0].left:
      lastListNode[0] = lastListNode[0].left
    return lastListNode[0]