• 分别用递归和非递归方式实现二叉树先序、中序和后序遍历


    【说明】:

      本文是左程云老师所著的《程序员面试代码指南》第三章中“分别用递归和非递归方式实现二叉树先序、中序和后序遍历”这一题目的C++复现。

      本文只包含问题描述、C++代码的实现以及简单的思路,不包含解析说明,具体的问题解析请参考原书。

      感谢左程云老师的支持。

    【题目】:

      用递归和非递归的方式,分别按照二叉树先序、中序和后序打印所有的节点。我们约定:先序遍历顺序为根、左、右;中序遍历顺序为左、根、右;后序遍历顺序为左、右、根。

     【思路】:

      解法:堆归的方式大家肯定都比较熟悉了。但是非递归的方式逻辑上还是蛮复杂的,从代码推敲推敲吧.........(去买本左老师的书也可以啊)^_^ ^_^ ^_^

    【编译环境】:

      CentOS6.7(x86_64)

      gcc 4.4.7

     【实现及测试】:

      声明代码:

     1 /*
     2  *文件名:bt_traversing.h
     3  *作者:
     4  *摘要:分别用递归和非递归的方式实现二叉树的先序、中序和后序遍历
     5  */
     6  
     7 #ifndef _BT_TRAVERSING
     8 #define _BT_TRAVERSING
     9 #include <iostream>
    10 using namespace std;
    11 
    12 class Node
    13 {
    14 public:
    15     Node(int data)
    16     {
    17         value = data;
    18         left = NULL;
    19         right = NULL;
    20     }
    21 public:
    22     int value;
    23     Node *left;
    24     Node *right;
    25 };
    26 
    27 //递归版
    28 void preOrderRecur(Node *head);
    29 void inOrderRecur(Node *head);
    30 void posOrderRecur(Node *head);
    31 
    32 //非递归版
    33 void preOrderUnRecur(Node *head);
    34 void inOrderUnRecur(Node *head);
    35 void posOrderUnRecur1(Node *head);
    36 void posOrderUnRecur2(Node *head);
    37 
    38 #endif
    View Code

      实现代码:

      1 /*
      2  *文件名:bt_traversing.cpp
      3  *作者:
      4  *摘要:分别用递归和非递归的方式实现二叉树的先序、中序和后序遍历
      5  */
      6  
      7 #include "bt_traversing.h"
      8 #include <iostream>
      9 #include <stack>
     10 
     11 using namespace std;
     12 
     13 void preOrderRecur(Node *head)
     14 {
     15     if(NULL == head)
     16         return ;
     17     cout << head->value << " ";
     18     preOrderRecur(head->left);
     19     preOrderRecur(head->right);
     20 }
     21 
     22 void inOrderRecur(Node *head)
     23 {
     24     if(NULL == head)
     25         return ;
     26     inOrderRecur(head->left);
     27     cout << head->value << " ";
     28     inOrderRecur(head->right);
     29 }
     30 
     31 void posOrderRecur(Node *head)
     32 {
     33     if(NULL == head)
     34         return ;
     35     posOrderRecur(head->left);
     36     posOrderRecur(head->right);
     37     cout << head->value << " ";
     38 }
     39 
     40 void preOrderUnRecur(Node *head)
     41 {
     42     if(NULL != head)
     43     {
     44         stack<Node*> s;
     45         s.push(head);
     46         while(!s.empty())
     47         {
     48             head = s.top();
     49             s.pop();
     50             cout << head->value << " ";
     51             if(NULL != head->right)
     52                 s.push(head->right);
     53             if(NULL != head->left)
     54                 s.push(head->left);
     55         }
     56     }
     57     cout << endl;
     58     return ;
     59 }
     60 
     61 void inOrderUnRecur(Node *head)
     62 {
     63     if(NULL != head)
     64     {
     65         stack<Node*> s;
     66         while(!s.empty() || NULL != head)
     67         {
     68             if(NULL != head)
     69             {
     70                 s.push(head);
     71                 head = head->left;
     72             }
     73             else
     74             {
     75                 head = s.top();
     76                 s.pop();
     77                 cout << head->value << " ";
     78                 head = head->right;
     79             }
     80         }
     81     }
     82     cout << endl;
     83     return ;
     84 }
     85 
     86 void posOrderUnRecur1(Node *head)
     87 {
     88     if(NULL != head)
     89     {
     90         stack<Node*> s1;
     91         stack<Node*> s2;
     92         s1.push(head);
     93         while(!s1.empty())
     94         {
     95             head = s1.top();
     96             s1.pop();
     97             s2.push(head);
     98             if(NULL != head->left)
     99                 s1.push(head->left);
    100             if(NULL != head->right)
    101                 s1.push(head->right);
    102         }
    103         while(!s2.empty())
    104         {
    105             cout << s2.top()->value << " ";
    106             s2.pop();
    107         }
    108     }
    109     cout << endl;
    110     return ;
    111 }
    112 
    113 void posOrderUnRecur2(Node *head)
    114 {
    115     if(NULL != head)
    116     {
    117         stack<Node*> s;
    118         s.push(head);
    119         Node *cur = NULL;
    120         while(!s.empty())
    121         {
    122             cur = s.top();
    123             if(NULL != cur->left && head != cur->left && head != cur->right)
    124                 s.push(cur->left);
    125             else if(NULL != cur->right && head != cur->right)
    126                 s.push(cur->right);
    127             else
    128             {
    129                 cout << s.top()->value << " ";
    130                 s.pop();
    131                 head = cur;
    132             }
    133         }
    134     }
    135     cout << endl;
    136     return ;    
    137 
    138 }
    View Code

      测试代码:

     1 /*
     2  *文件名:main.cpp
     3  *作者:
     4  *摘要:测试代码
     5  */
     6 
     7 #include "bt_traversing.h"
     8 #include <iostream>
     9 using namespace std;
    10 
    11 //由数组创建完全二叉树
    12 void createBT(Node **head,int arr[],int len,int index=0)
    13 {
    14     if(index > len-1)
    15         return;
    16     (*head) = new Node(arr[index]);
    17     createBT(&((*head)->left),arr,len,2*index+1);
    18     createBT(&((*head)->right),arr,len,2*index+2);
    19 }
    20 
    21 int main()
    22 {
    23     int arr[] = {1,2,3,4,5,6,7,8,9};
    24     Node *head = NULL;
    25     createBT(&head,arr,9);
    26 
    27     cout << "Using recursive method to achieve pre-order traversal:" << endl;
    28     preOrderRecur(head);
    29     cout << endl;
    30     cout << "Using recursive method to achieve in-order traversal:" << endl;
    31     inOrderRecur(head);
    32     cout << endl;
    33     cout << "Using recursive method to achieve pos-order traversal:" << endl;
    34     posOrderRecur(head);
    35     cout << endl;
    36     
    37     cout << "Without using recursive way to achieve the pre-order traversal:"<< endl;
    38     preOrderUnRecur(head);
    39     cout << "Without using recursive way to achieve the in-order traversal:"<< endl;
    40     inOrderUnRecur(head);
    41     cout << "The first method achieve the pos-order traversal without using recursive:"<< endl;
    42     posOrderUnRecur1(head);
    43     cout << "The second method achieve the pos-order traversal without using recursive:"<< endl;
    44     posOrderUnRecur2(head);
    45     return 0;
    46 }
    View Code

    注:

      转载请注明出处;

      转载请注明源思路来自于左程云老师的《程序员代码面试指南》。

  • 相关阅读:
    CentOS8安装Mysql5.7
    CentOS8搭建FTP服务器
    CentOS8安装jdk1.8
    基于可穿戴设备的医疗监护系统
    air530GPS数据通过air202上传阿里云
    bzoj2594: [Wc2006]水管局长数据加强版
    bzoj3091: 城市旅行
    Problem A. Array Factory XVII Open Cup named after E.V. Pankratiev Stage 4: Grand Prix of SPb, Sunday, Octorber 9, 2016
    hdu5716
    bzoj2002: [Hnoi2010]Bounce 弹飞绵羊
  • 原文地址:https://www.cnblogs.com/PrimeLife/p/5457320.html
Copyright © 2020-2023  润新知