python数据结构之二叉树的遍历实例

有次序：
　NLR、LNR、LRN

遍历的命名

　根据访问结点操作发生位置命名：
NLR：前序遍历(PreorderTraversal亦称(先序遍历)) ——访问结点的操作发生在遍历其左右子树之前。
LNR：中序遍历(InorderTraversal) ——访问结点的操作发生在遍历其左右子树之中(间)。
LRN：后序遍历(PostorderTraversal) ——访问结点的操作发生在遍历其左右子树之后。

注：由于被访问的结点必是某子树的根，所以N(Node)、L(Left subtlee)和R(Right subtree)又可解释为根、根的左子树和根的右子树。NLR、LNR和LRN分别又称为先根遍历、中根遍历和后根遍历。

遍历算法

1)．先序遍历的递归算法定义：
若二叉树非空，则依次执行如下操作：
a.访问根结点
b.遍历左子树
c.遍历右子树

2)．中序遍历的递归算法定义：
若二叉树非空，则依次执行如下操作：
a.遍历左子树
b.访问根结点
c.遍历右子树

3)．后序遍历得递归算法定义：
若二叉树非空，则依次执行如下操作：
a.遍历左子树
b.遍历右子树
c.访问根结点

一、二叉树的递归遍历：

代码如下:

# -*- coding: utf - 8 - *-

class TreeNode(object):

def __init__(self, left=0, right=0, data=0):
self.left = left
self.right = right
self.data = data


class BTree(object):

def __init__(self, root=0):
self.root = root

def is_empty(self):
if self.root is 0:
return True
else:
return False

def preorder(self, treenode):
'前序（pre-order，NLR）遍历'
if treenode is 0:
return
print treenode.data
self.preorder(treenode.left)
self.preorder(treenode.right)

def inorder(self, treenode):
'中序（in-order，LNR'
if treenode is 0:
return
self.inorder(treenode.left)
print treenode.data
self.inorder(treenode.right)

def postorder(self, treenode):
'后序（post-order，LRN）遍历'
if treenode is 0:
return
self.postorder(treenode.left)
self.postorder(treenode.right)
print treenode.data


node1 = TreeNode(data=1)
node2 = TreeNode(node1, 0, 2)
node3 = TreeNode(data=3)
node4 = TreeNode(data=4)
node5 = TreeNode(node3, node4, 5)
node6 = TreeNode(node2, node5, 6)
node7 = TreeNode(node6, 0, 7)
node8 = TreeNode(data=8)
root = TreeNode(node7, node8, 'root')

bt = BTree(root)

print u'''

#生成的二叉树

# ------------------------
# root
# 7 8
# 6
# 2 5
# 1 3 4
#
# -------------------------

'''
print '前序（pre-order，NLR）遍历 ：\n'
bt.preorder(bt.root)

print '中序（in-order，LNR) 遍历 ：\n'
bt.inorder(bt.root)

print '后序（post-order，LRN）遍历 ：\n'
bt.postorder(bt.root)

二、.二叉树的非递归遍历

下面就用非递归的方式实现一遍。主要用到了 stack 和 queue维护一些数据节点：

代码如下:

# -*- coding: utf - 8 - *-


class TreeNode(object):

def __init__(self, left=0, right=0, data=0):
self.left = left
self.right = right
self.data = data


class BTree(object):

def __init__(self, root=0):
self.root = root

def is_empty(self):
if self.root is 0:
return True
else:
return False

def preorder(self, treenode):
'前序（pre-order，NLR）遍历'
stack = []
while treenode or stack:
if treenode is not 0:
print treenode.data
stack.append(treenode)
treenode = treenode.left
else:
treenode = stack.pop()
treenode = treenode.right

def inorder(self, treenode):
'中序（in-order，LNR) 遍历'
stack = []
while treenode or stack:
if treenode:
stack.append(treenode)
treenode = treenode.left
else:
treenode = stack.pop()
print treenode.data
treenode = treenode.right

# def postorder(self, treenode):
# stack = []
# pre = 0
# while treenode or stack:
# if treenode:
# stack.append(treenode)
# treenode = treenode.left
# elif stack[-1].right != pre:
# treenode = stack[-1].right
# pre = 0
# else:
# pre = stack.pop()
# print pre.data

def postorder(self, treenode):
'后序（post-order，LRN）遍历'
stack = []
queue = []
queue.append(treenode)
while queue:
treenode = queue.pop()
if treenode.left:
queue.append(treenode.left)
if treenode.right:
queue.append(treenode.right)
stack.append(treenode)
while stack:
print stack.pop().data

def levelorder(self, treenode):
from collections import deque
if not treenode:
return
q = deque([treenode])
while q:
treenode = q.popleft()
print treenode.data
if treenode.left:
q.append(treenode.left)
if treenode.right:
q.append(treenode.right)


node1 = TreeNode(data=1)
node2 = TreeNode(node1, 0, 2)
node3 = TreeNode(data=3)
node4 = TreeNode(data=4)
node5 = TreeNode(node3, node4, 5)
node6 = TreeNode(node2, node5, 6)
node7 = TreeNode(node6, 0, 7)
node8 = TreeNode(data=8)
root = TreeNode(node7, node8, 'root')


bt = BTree(root)

print u'''

#生成的二叉树

# ------------------------
# root
# 7 8
# 6
# 2 5
# 1 3 4
#
# -------------------------

'''
print '前序（pre-order，NLR）遍历 ：\n'
bt.preorder(bt.root)

print '中序（in-order，LNR) 遍历 ：\n'
bt.inorder(bt.root)

print '后序（post-order，LRN）遍历 ：\n'
bt.postorder(bt.root)

print '层序（level-order，LRN）遍历 ：\n'
bt.levelorder(bt.root)