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本文实例讲述了Python实现重建二叉树的三种方法。分享给大家供大家参考，具体如下：

学习算法中，探寻重建二叉树的方法：

用input 前序遍历顺序输入字符重建
前序遍历顺序字符串递归解析重建
前序遍历顺序字符串堆栈解析重建

如果懒得去看后面的内容，可以直接点击此处本站下载完整实例代码。

思路

学习算法中，python 算法方面的资料相对较少，二叉树解析重建更少，只能摸着石头过河。

通过不同方式遍历二叉树，可以得出不同节点的排序。那么，在已知节点排序的前提下，通过某种遍历方式，可以将排序进行解析，从而构建二叉树。

应用上来将，可以用来解析多项式、可以解析网页、xml等。

本文采用前序遍历方式的排列，对已知字符串进行解析，并生成二叉树。新手，以解题为目的，暂未优化，未能体现 Python 简洁、优美。请大牛不吝指正。

首先采用 input 输入

节点类

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

input 方法重建二叉树

 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

以下两种方法，使用预先编好的字符串，通过 list 方法转换为 list 传入进行解析

myTree 为实例化一个空树

调用递归方法重建二叉树

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

递归方法重建二叉树

 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

调用堆栈方法重建二叉树

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

使用堆栈重建二叉树

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

显示二叉树

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)

打印二叉树的代码，采用某仁兄代码，在此感谢。

input 输入及显示二叉树结果