//哈夫曼樹
//楊鑫
#include <stdio.h>
#include <stdlib.h>
typedef int ElemType;
struct BTreeNode
{
    ElemType data;
    struct BTreeNode* left;
    struct BTreeNode* right;
};

//遍歷哈夫曼樹
void PrintBTree_int(struct BTreeNode* BT)
{
    if (BT != NULL)
    {
        printf("%d", BT->data); 
        if (BT->left != NULL || BT->right != NULL)
        {
            printf(" ( ");
            PrintBTree_int(BT->left); //輸出左子樹
            if (BT->right != NULL)
                printf(" , ");
            PrintBTree_int(BT->right); //輸出右子樹
            printf(" ) ");
        }
    }
}

//建立哈夫曼樹
struct BTreeNode* CreateHuffman(ElemType a[], int n)
{
    int i, j;
    struct BTreeNode **b, *q;
    b = malloc(n*sizeof(struct BTreeNode));
    for (i = 0; i < n; i++) 							//動態記憶體分配
    {
        b[i] = malloc(sizeof(struct BTreeNode));
        b[i]->data = a[i];
        b[i]->left = b[i]->right = NULL;
    }
    for (i = 1; i < n; i++)
    {
        //k1表示森林中具有最小權值的樹根結點的下標，k2為次最小的下標
        int k1 = -1, k2;
        for (j = 0; j < n; j++)						//讓k1初始指向森林中第一棵樹，k2指向第二棵
        {
            if (b[j] != NULL && k1 == -1)
            {
                k1 = j;
                continue;
            }
            if (b[j] != NULL)
            {
                k2 = j;
                break;
            }
        }
        for (j = k2; j < n; j++)					//構造最優解
        {
            if (b[j] != NULL)
            {
                if (b[j]->data < b[k1]->data)
                {
                    k2 = k1;
                    k1 = j;
                }
                else if (b[j]->data < b[k2]->data)
                    k2 = j;
            }
        }
        q = malloc(sizeof(struct BTreeNode));
        q->data = b[k1]->data + b[k2]->data;
        q->left = b[k1];
        q->right = b[k2];

        b[k1] = q;
        b[k2] = NULL;
    }
    free(b); 
    return q; 
}

//計算帶權路徑
ElemType WeightPathLength(struct BTreeNode* FBT, int len)//len初始為0
{
    if (FBT == NULL) //空樹返回0
        return 0;
    else
    {
        if (FBT->left == NULL && FBT->right == NULL)
            return FBT->data * len;
        else 		
            return WeightPathLength(FBT->left,len+1)+WeightPathLength(FBT->right,len+1);
    }
}

//構造哈夫曼編碼
void HuffManCoding(struct BTreeNode* FBT, int len)
{
    static int a[10];
    if (FBT != NULL)    
	{
        if (FBT->left == NULL && FBT->right == NULL)
        {
            int i;
            printf("結點的值為%d的編碼：", FBT->data);
            for (i = 0; i < len; i++)
                printf("%d", a[i]);
            printf("\n");
        }
        else
        {   
            a[len] = 0;
            HuffManCoding(FBT->left, len + 1);
            a[len] = 1;
            HuffManCoding(FBT->right, len + 1);
        }
    }
}


int main()
{
    int n, i;
    ElemType* a;
    struct BTreeNode* fbt;
    printf("請輸入帶權結點數n：\n");
    while(1)
    {
        scanf("%d", &n);
        if (n > 1)
            break;
        else
            printf("重輸n值：");
    }
    a = malloc(n * sizeof(ElemType));
    printf("輸入%d個整數作為權值：\n", n);
    for (i = 0; i < n; i++)
        scanf(" %d", &a[i]);
    fbt = CreateHuffman(a, n);
    printf("哈夫曼樹如下：\n");
    PrintBTree_int(fbt);
    printf("\n");
    printf("哈夫曼樹的帶權路徑長度：\n");
    printf("%d\n", WeightPathLength(fbt, 0));
    printf("樹中每個葉子結點的哈夫曼編碼：\n");
    HuffManCoding(fbt, 0);
	return 0;
}

 

如圖：