// 构造赫夫曼树
#include <iostream>
using namespace std;
//哈夫曼树的存储表示
typedef struct
{
int weight; //节点的权值
int parent, lchild, rchild; //节点的双亲,左孩子和右孩子
} HTNode, *HuffmanTree;
//查处权值最小且双亲为0的2的节点
void Select(HuffmanTree HT, int len, int &s1, int &s2)
{
int i, min1 = 0x3f3f3f3f, min2 = 0x3f3f3f3f; //先赋予最大值
for (i = 1; i <= len; i++)
{
if (HT[i].weight < min1 && HT[i].parent == 0)
{
min1 = HT[i].weight;
s1 = i;
}
}
int temp = HT[s1].weight; //将原值存放起来,然后先赋予最大值,防止s1被重复选择
HT[s1].weight = 0x3f3f3f3f;
for (i = 1; i <= len; i++)
{
if (HT[i].weight < min2 && HT[i].parent == 0)
{
min2 = HT[i].weight;
s2 = i;
}
}
HT[s1].weight = temp; //恢复原来的值
}
void CreatHuffmanTree(HuffmanTree &HT, int n)
{
//构造赫夫曼树HT
int m, s1, s2, i;
if (n <= 1)
return;
m = 2 * n - 1;
HT = new HTNode[m + 1]; //0号单元未用,所以需要动态分配m+1个单元,HT[m]表示根结点
for (i = 1; i <= m; ++i) //将1~m号单元中的双亲、左孩子,右孩子的下标都初始化为0
{
HT[i].parent = 0;
HT[i].lchild = 0;
HT[i].rchild = 0;
}
cout << "请输入叶子结点的权值:\n";
for (i = 1; i <= n; ++i) //输入前n个单元中叶子结点的权值
cin >> HT[i].weight;
/*――――――――――初始化工作结束,下面开始创建赫夫曼树――――――――――*/
for (i = n + 1; i <= m; ++i)
{ //通过n-1次的选择、删除、合并来创建赫夫曼树
Select(HT, i - 1, s1, s2);
//在HT[k](1≤k≤i-1)中选择两个其双亲域为0且权值最小的结点,
// 并返回它们在HT中的序号s1和s2
HT[s1].parent = i;
HT[s2].parent = i;
//得到新结点i,从森林中删除s1,s2,将s1和s2的双亲域由0改为i
HT[i].lchild = s1;
HT[i].rchild = s2; //s1,s2分别作为i的左右孩子
HT[i].weight = HT[s1].weight + HT[s2].weight; //i 的权值为左右孩子权值之和
} //for
} // CreatHuffmanTree
int main()
{
HuffmanTree HT;
int n;
cout << "请输入叶子结点的个数:\n";
cin >> n;
CreatHuffmanTree(HT, n);
cout << "哈夫曼树建立完毕!\n";
}
哈夫曼树的创建(详细步骤)
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转载自blog.csdn.net/qq_47142993/article/details/107769751
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