Python实现重建二叉树的三种方法详解

class treeNode:
 def __init__(self, rootObj = None, leftChild = None, rightChild = None):
  self.key = rootObj
  self.leftChild = None
  self.rightChild = None

 def createTreeByInput(self, root):
  tmpKey = raw_input("please input a key, input '#' for Null")
  if tmpKey == '#':
   root = None
  else:
   root = treeNode(rootObj=tmpKey)
   root.leftChild = self.createTreeByInput(root.leftChild)
   root.rightChild = self.createTreeByInput(root.rightChild)
  return root

 treeElementList = '124#8##5##369###7##'
 myTree = myTree.createTreeByListWithRecursion(list(treeElementList))
 printBTree(myTree, 0)

 def createTreeByListWithRecursion(self, preOrderList):
  """
  根据前序列表重建二叉树
  :param preOrder: 输入前序列表
  :return: 二叉树
  """
  preOrder = preOrderList
  if preOrder is None or len(preOrder) <= 0:
   return None
  currentItem = preOrder.pop(0) # 模拟C语言指针移动
  if currentItem is '#':
   root = None
  else:
   root = treeNode(currentItem)
   root.leftChild = self.createTreeByListWithRecursion(preOrder)
   root.rightChild = self.createTreeByListWithRecursion(preOrder)
  return root

 treeElementList = '124#8##5##369###7##'
 myTree = myTree.createTreeByListWithStack(list(treeElementList))
 printBTree(myTree, 0)

def createTreeByListWithStack(self, preOrderList):
 """
 根据前序列表重建二叉树
 :param preOrder: 输入前序列表
 :return: 二叉树
 """
 preOrder = preOrderList
 pStack = SStack()
 # check
 if preOrder is None or len(preOrder) <= 0 or preOrder[0] is '#':
  return None
 # get the root
 tmpItem = preOrder.pop(0)
 root = treeNode(tmpItem)
 # push root
 pStack.push(root)
 currentRoot = root
 while preOrder:
  # get another item
  tmpItem = preOrder.pop(0)
  # has child
  if tmpItem is not '#':
   # does not has left child, insert one
   if currentRoot.leftChild is None:
    currentRoot = self.insertLeft(currentRoot, tmpItem)
    pStack.push(currentRoot.leftChild)
    currentRoot = currentRoot.leftChild
   # otherwise insert right child
   elif currentRoot.rightChild is None:
    currentRoot = self.insertRight(currentRoot, tmpItem)
    pStack.push(currentRoot.rightChild)
    currentRoot = currentRoot.rightChild
  # one child is null
  else:
   # if has no left child
   if currentRoot.leftChild is None:
    currentRoot.leftChild = None
    # get another item fill right child
    tmpItem = preOrder.pop(0)
    # has right child
    if tmpItem is not '#':
     currentRoot = self.insertRight(currentRoot, tmpItem)
     pStack.push(currentRoot.rightChild)
     currentRoot = currentRoot.rightChild
    # right child is null
    else:
     currentRoot.rightChild = None
     # pop itself
     parent = pStack.pop()
     # pos parent
     if not pStack.is_empty():
      parent = pStack.pop()
     # parent become current root
     currentRoot = parent
     # return from right child, so the parent has right child, go to parent's parent
     if currentRoot.rightChild is not None:
      if not pStack.is_empty():
       parent = pStack.pop()
       currentRoot = parent
   # there is a leftchild ,fill right child with null and return to parent
   else:
    currentRoot.rightChild = None
    # pop itself
    parent = pStack.pop()
    if not pStack.is_empty():
     parent = pStack.pop()
    currentRoot = parent
 return root

def printBTree(bt, depth):
 '''''
 递归打印这棵二叉树，#号表示该节点为NULL
 '''
 ch = bt.key if bt else '#'
 if depth > 0:
  print '%s%s%s' % ((depth - 1) * ' ', '--', ch)
 else:
  print ch
 if not bt:
  return
 printBTree(bt.leftChild, depth + 1)
 printBTree(bt.rightChild, depth + 1)