python数据结构之二叉树的遍历实例

2019-11-25 18:25:40

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遍历方案
   　从二叉树的递归定义可知，一棵非空的二叉树由根结点及左、右子树这三个基本部分组成。因此，在任一给定结点上，可以按某种次序执行三个操作：
   　1).访问结点本身(N)
   　2).遍历该结点的左子树(L)
   　3).遍历该结点的右子树(R)

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：

复制代码代码如下:

# -*- 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：

复制代码代码如下:

# -*- 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)