Python--实现二叉树的遍历操作

一、首先二叉树的定义：

            
              class TreeNode:
  def __init__(self, x):
    self.val = x
    self.left = None
    self.right = None
            
          

        构建一棵二叉树：

            
              class Node(object):
    def __init__(self, val):
        self.val = val
        self.lchild = None
        self.rchild = None


class Tree(object):

    def __init__(self):
        self.root = None
        self.myQueue = []

    def add(self, val):
        node = Node(val)

        if self.root == None:
            self.root = node
            self.myQueue.append(self.root)
        else:
            while True:
                point = self.myQueue[0]

                if point.lchild == None:
                    point.lchild = node
                    self.myQueue.append(point.lchild)
                    return
                elif point.rchild == None:
                    point.rchild = node
                    self.myQueue.append(point.rchild)
                    self.myQueue.pop(0)
                    return

            
          

 

二、前序遍历

            
              递归实现

def preorder(root,res=[]):
  if not root:
    return 
  res.append(root.val)
  preorder(root.lchild,res)
  preorder(root.rchild,res)
  return res

迭代实现

def preorder(root):
  res = []
  if not root: 
    return []
  stack = [root]
  while stack:
    node = stack.pop()
    res.append(node.val)
    if node.rchild:
      stack.append(node.rchild)
    if node.lchild:
      stack.append(node,lchild)
  return res
            
          

  三、中序遍历

            
              递归实现

def inorder(root,res=[]):
  if not root:
    return 
  inorder(root.lchild,res)
  res.append(root.val)
  inorder(root.rchild,res)
  return res


迭代实现

def inorder(root):
  stack = []
  node = root
  res = []
  while stack or node:
    while node:
      stack.append(node)
      node = node.lchild
    node = stack.pop()
    res.append(node.val)
    node = node.rchild
  return res

            
          

  四、后序遍历

            
              递归实现

def laorder(root,res=[]):
  if not root:
    return 
  laorder(root.lchild,res)
  laorder(root.rchild,res)
  res.append(root.val)
  return res



迭代实现

def laorder(root):
  stack = [root]
  res = []
  while stack:
    node = stack.pop()
    if node.lchild:
      stack.append(node.left)
    if node.rchild:
      stack.append(node.right)
    res.append(node.val)
  return res[::-1]
            
          

五、层次遍历 

            
              迭代

def levelorder(root):
  queue = [root]
  res = []
  while queue:
    node = queue.pop(0)
    if node.lchild: 
      queue.append(node.lchild)
    if node.rchild:
      queue.append(node.rchild)
    res.append(node.val)
  return res
            
          

 运行结果：

 

 参考博客：https://www.jb51.net/article/157422.htm、https://blog.csdn.net/Bone_ACE/article/details/46718683、https://blog.csdn.net/wzngzaixiaomantou/article/details/81294915