标签：哈夫曼树 数据结构 二叉树 知识原理 算法理解 全局变量
关于哈夫曼树其基本定义我就在这不多说了，我写的这个哈夫曼树没有太多的功能，就是我们参考书上的一个作业，就俩个功能，一个是表示孩子双亲表示法，，还有就是吧这个最有二叉树给画出来，其他的就没有了。另外在今天我调试的时候发现我的代码还是有很多欠缺的地方的，比如哦说我的画图功能中就没有办法完整的输出，而是无法输出最后四个结点。原因我还在寻找，
下面就哈夫曼树我来谈谈我当时的构思：
首先我是吧他当作一个二叉树来写的。我首先申明一个结点构造的结点函数element，在element中我全部都是用的int 来定义的，因为在哈发满树中我们需要的是每个节点的权值，所以我们就不需要用到类模板 了，而是直接用int来解决这个类。然偶就是在这个类中添砖加瓦的事情了。自己想要什么函数，在里面申明解释就行了。我在构造函数的时候其中有selected和画图函数比较难搞。下面#ifndef HFMTREE_H
#define HFMTREE_H
const int Max = 100;
#include"BiTree.h"
#include<iostream>
using namespace std;
struct element
{
int weight;
int lchild, rchild, parent;
};
class HfmTree
{
public:
HfmTree(int size);
void Creat(int n);
void SeLect(int k, int &s1, int &s2 ,int z);
void reset(int size);
void print(int size);
void Makeary(int size);
void printInl(int size, int i);
void printInr(int size, int i);
void Build(int size);
private:
int m1;
int top = 0;
element *root;
int maxsize;
int *pz;
element *root3;
};
void HfmTree::reset(int size)
{
int m;
m = 2 * size - 1;
pz = new int[m];
int i;
pz[0] = root[m].weight;
for (i = 1; i < m; i++)
pz[i] = 0;
}
void HfmTree::Build(int size)
{
BiTree<int> zxh1(pz, 2 * size - 1, 0);
cout << zxh1;
}
void HfmTree::printInl(int m, int i)
{
if (m == -1)return;
if (m!=-1)
{
//pz[i - 1] = root[m].weight;
pz[2 * i - 1] = root[root[m].lchild].weight;
pz[2 * i] = root[root[m].rchild].weight;
m = root[m].lchild;
printInl(m, 2 * i);
m = root[m].lchild;
}
}
void HfmTree::printInr(int m, int i)
{
if (m == -1)return;
if (m!=-1)
{
//pz[i - 1] = root[m].weight;
pz[2 * i - 1] = root[root[m].lchild].weight;
pz[2 * i] = root[root[m].rchild].weight;
m = root[m].rchild;
printInr(m, 2*i+1);
m = root[m].rchild;
}
}
void HfmTree::Makeary(int size)
{
int m;
m = 2 * size - 1;
pz = new int[m];
int i;
for (i = 1; i <= m; i++)
pz[i] = 0;
i = 1;
while (root[m].parent != size-1)
{
if (m == -1)break;
pz[i-1] = root[m].weight;
i=2*i;
m = root[m].lchild;
}
m = 2 * size - 1; i = 1;
while (root[m].parent != size-1)
{
if (m == -1)break;
pz[i-1] = root[m].weight;
i = 2 * i + 1;
m = root[m].rchild;
}
BiTree<int> zxh1(pz, 2*size+1, 0);
cout << zxh1;
}
void HfmTree::SeLect(int k, int &s1, int &s2 ,int z)
{
int w1, w2, i = 1,j;
w1 = w2 = Max;
//s1 = s2 = i;
for (i = 1; i <= k; i++)
{
//if (i == s1 || i == s2)continue;
if (root[i].parent==-1)
{
if (root[i].weight < w1)
{
w2 = w1;
w1 = root[i].weight;
s2 = s1;
s1 = i; //w1,s1是对应的，且他们的weight是较小者
}
else if (root[i].weight < w2)
{
w2 = root[i].weight;
s2 = i;
}
}
}
}
void HfmTree::Creat(int n)
{
int *w1;
int i,m,*w,s1,s2; element *p;
if (n < 1)throw"构建有误！";
w = new int[2 * n - 1];
w1 = new int[2 * n - 1];
m = 2 * n - 1; //n个结点过程中有2*n-1个结点
m1 = m;
root = new element[m + 1];
for (i =1; i <= 2 * n - 1; i++)
{
//root[i].weight = NULL;
root[i].lchild = -1;
root[i].rchild = -1;
root[i].parent = -1;
}
for (p = root + 1, i = 1; i <= n; i++, p++)
{
cout << "请输入全职:" << endl;
cin >> w[i];
p->weight = w[i];
p->lchild = -1;
p->rchild = -1;
p->parent = -1;
}
for (i = n+1 ; i <= m; i++, p++)
{
p->weight = NULL;
p->lchild = -1;
p->rchild = -1;
p->parent = -1;
}
for (i = n+1 ; i <= m; i++)
{
SeLect(i-1 , s1, s2,m);
root[s1].parent = i-1;
root[s2].parent = i-1;
root[i].weight = root[s1].weight + root[s2].weight;
root[i].lchild = s1;
root[i].rchild = s2;
}
for (int i = 1; i <= n; i++)
{
w1[i-1] = w[n - i + 1];
}
}
HfmTree::HfmTree(int size)
{
Creat(size);
}
void HfmTree::print(int size)
{
int m, i;
m = 2 * size - 1;//树结点总个数。
cout << "构建的哈夫曼树为" << endl;
cout << "no" <<" "<< "weight" <<" "<<"Parent"<<" "<< "lchild" <<" "<< "rchild" << endl;
for (i = 1; i <= m; i++){
cout << i <<" "<< root[i].weight <<" "<<root[i].parent<<" "<< root[i].lchild <<" "<< root[i].rchild << endl;
}
}
#endif
——————————————————————————————————————————————————————————————————————
#include"HfmTree.h"
//#include"BiTree.h"
#include<iostream>
using namespace std;
void main()
{
int num;
cout << "请输入您想构造的哈夫曼树的结点个数(num):"; cin >> num;
HfmTree zxh(num);
zxh.print(num);
//zxh.Makeary(num);
zxh.reset(num);
zxh.printInl(2*num-1,1);
zxh.printInr(2*num-1, 1);
zxh.Build(num);
}主要用代码解释
标签：哈夫曼树 数据结构 二叉树 知识原理 算法理解 全局变量
原文地址：http://blog.csdn.net/u013132051/article/details/40934287
