本文包含两个文件的代码和一张测试效果图：

• HuffmanCD.h文件： 从叶到根逆向求哈夫曼编码
• HuffmanCodingTest.cpp文件： 用于测试
• 效果图：（如下）

效果图：

HuffmanCD.h文件：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
 
typedef struct{
	int weight;
	int parent,lchild,rchild;
}HTNode,*HuffmanTree;
 
typedef char** HuffmanCode; //动态分配数组存储哈夫曼编码表

void Select(HuffmanTree HT, int n, int &s1, int &s2)
{
    int minum;      // 定义一个临时变量保存最小值?
    for(int i=1; i<=n; i++)     // 以下是找到第一个最小值
    {
        if(HT[i].parent == 0)
        {
            minum = i;
            break;
        }
    }
    for(int i=1; i<=n; i++)
    {
        if(HT[i].parent == 0)
            if(HT[i].weight < HT[minum].weight)
                minum = i;
    }
    s1 = minum;
    // 以下是找到第二个最小值，且与第一个不同
    for(int i=1; i<=n; i++)     
    {
        if(HT[i].parent == 0 && i != s1)
        {
            minum = i;
            break;
        }
    }
    for(int i=1; i<=n; i++)
    {
        if(HT[i].parent == 0 && i != s1)
            if(HT[i].weight < HT[minum].weight)
                minum = i;
    }
    s2 = minum;
}

void HuffmanInit(HuffmanTree &HT, HuffmanCode &HC, char* ch, int* w, int n){

	//ch存放n个字符，w存放n个字符的权值，构造哈夫曼树HT，并求出n个字符的哈夫曼编码HC
	int m;	
	int i; 
	char* cd; //临时存储每个字符的编码串
	int start; //记录编码在cd中存放的位置，初始指向最后，以便于逆向求编码 
	int c; //记录当前待编码字符的下标i
	int f; //记录前待编码字符的下标i的双亲结点的下标 
	int s1, s2; //记录权值最小两个编码对应下标 
	HuffmanTree p; 
	if(n <= 1){
		return;
	}
	m = 2 * n - 1; //n个叶子结点的哈夫曼树共有2n-1个结点 
	HT = (HuffmanTree)malloc((m + 1) * sizeof(HTNode)); //为了表示方便，0号单元未用
	for(p = HT + 1, i = 1; i <= n; i++, p++, w++){
		(*p).weight = *w;
		(*p).parent = 0;
		(*p).lchild = 0;
		(*p).rchild = 0;
	} 
	for(; i <= m; i++, p++){
		
		(*p).weight = 0;
		(*p).parent = 0;
		(*p).lchild = 0;
		(*p).rchild = 0;
	}
	for(i = n + 1; i <= m; i++){ //建哈夫曼树 
		Select(HT, i - 1, s1, s2);
		HT[s1].parent = i;
		HT[s2].parent = i;
		HT[i].lchild = s1;
		HT[i].rchild = s2;
		HT[i].weight = HT[s1].weight + HT[s2].weight;
	}
	//从叶到根逆向求每个字符的哈夫曼编码
	HC = (HuffmanCode)malloc((n + 1) * sizeof(char*)); //分配n个字符编码的头指针
	cd = (char*)malloc(n * sizeof(char)); //分配求编码的工作空间
	cd[n-1] = '\0'; 
	for(i = 1; i <= n; i++){ //求解n个字符的编码，循环n次 
		start = n - 1;
		for(c = i, f = HT[i].parent; f != 0; c = f, f = HT[f].parent){ //从叶到根逆向求编码 
			if(HT[f].lchild == c){
				HT[c].weight = 0;
				cd[--start] = '0';
			}
			else{
				HT[c].weight = 1;
				cd[--start] = '1';
			}
		}
		HC[i] = (char*)malloc((n - start) * sizeof(char)); //为第i个字符编码分配空间
		strcpy(HC[i], &cd[start]); //从cd复制编码串到HC 
		printf("%c: ", ch[i-1]);
		puts(HC[i]);
	} 
} 

void HuffmanCoding(HuffmanTree HT, HuffmanCode HC, char* input_ch, char* ch, int n){
	int i, j;
	int len = strlen(input_ch);
	printf("%s经哈夫曼编码后为: ", input_ch);
	for(i = 0; i < len; i++){
		for(j = 0; j < n; j++){
			if(input_ch[i] == ch[j]){
				printf("%s", HC[j+1]);
			    break;
			}
		}
	}
}

//解码并未完全代码实现出来，只稍微写了点 
//void HuffmanDeCoding(HuffmanTree HT, HuffmanCode HC, char* ch_num, int n){
//	int i;
//	int len = strlen(ch_num);
//	HuffmanTree p = HT + 1;
//	char cd[101];
//	int count = 0;
//	for(i = 0; i < len; i++){
//		if(ch_num[i] == '0' && (*p).lchild != 0){
//			(*p) = (*p).lchild;
//			cd[count] = ch_num[i];
//			count++;
//		}
//		else if(ch_num[i] == '1' && *(p)->rchild != 0){
//			(*p) = (*p).rchild;
//			cd[count] = ch_num[i];
//			count++;
//		}
//		else{
//			puts(cd);
//			p = HT + 1;	
//			char cd[101];
//			int count = 0;
//		}
//	}
//}

HuffmanCodingTest.cpp文件：

#include "HuffmanCD.h"

int main()
{
    HuffmanTree HT;
    char** HC;
    int n = 6, wei;
    char ch[6] = {'A', 'B', 'C', 'D', 'E', 'F'}; 
    int w[6] = {27, 8, 15, 15, 30, 5};
    
    HuffmanInit(HT, HC, ch, w, n);
    printf("\n");
    
    char input_ch[101] = "BADCADFEED";
    HuffmanCoding(HT, HC, input_ch, ch, n);
    printf("\n");
    
//    char input_num[101] = "101100101"; //CBA
//    HuffmanDeCoding(HT, HC, input_num, n);
}