数据结构学习-二叉树的基本运算

问题描述：

构造一棵哈夫曼树，输出对应的哈夫曼编码和平均查找长度。并用表7.8所示的数据进行验证。
表7.8 单词及出现的频度

单词 The of a to and in that he is at on for His are be 出现频度 1192 677 541 518 462 450 242 195 190 181 174 157 138 124 123

---

代码：

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;const int n=15;typedef struct{    char data[5];    double weight;    int parent;    int lchild;    int rchild;}HTNode;typedef struct{    char cd[n];    int start;}HCode;void CreateHT(HTNode ht[]){    int i,k,lnode,rnode;    double min1,min2;    for(i=0;i<2*n-1;i++)        ht[i].lchild=ht[i].rchild=ht[i].parent=-1;    for(i=n;i<2*n-1;i++)    {        min1=min2=9999;        lnode=rnode=-1;        for(k=0;k<i;k++)        {            if(ht[k].parent==-1)            {                if(ht[k].weight<min1)                {                    min2=min1;rnode=lnode;                    min1=ht[k].weight;lnode=k;                }                else if(ht[k].weight<min2)                {                    min2=ht[k].weight;rnode=k;                }            }        }        ht[i].weight=ht[lnode].weight+ht[rnode].weight;        ht[i].lchild=lnode;ht[i].rchild=rnode;        ht[lnode].parent=i;ht[rnode].parent=i;    }}void CreatHCode(HTNode ht[],HCode hcd[]){    int i,f,c;    HCode hc;    for(i=0;i<n;i++)    {        hc.start=n;        c=i;        f=ht[i].parent;        while(f!=-1)        {            if(ht[f].lchild==c)                hc.cd[--hc.start]='0';            else                hc.cd[--hc.start]='1';            c=f;f=ht[f].parent;        }        hcd[i]=hc;    }}void DispHCode(HTNode ht[],HCode hc[]){    int i,j,k;    double sum=0,m=0;    cout<<"输出哈弗曼编码：\n";    for(i=0;i<n;i++)    {        j=0;        cout<<"\t"<<ht[i].data<<":\t";        for(k=hc[i].start;k<n;k++)        {            cout<<hc[i].cd[k];            j++;        }        cout<<endl;        m+=ht[i].weight;        sum+=ht[i].weight*j;    }    cout<<"平均长度="<<sum/m<<endl;}int main(){    int w[]={1192,677,541,518,462,450,242,195,190,181,174,157,138,124,123};    char *str[]={"The","of","a","to","and","in","that","he","is","at","on","for","His","are","be"};    HTNode ht[2*n-1];    HCode hc[n];    for(int i=0;i<n;i++)    {        strcpy(ht[i].data,str[i]);        ht[i].weight=w[i];    }    CreateHT(ht);    CreatHCode(ht,hc);    DispHCode(ht,hc);    return 0;}

---

运行结果：