问题描述：有一电文共使用五种字符A,B,C,D,E,其出现频率依次为4,7,5,2,9。构造对应的编码哈夫曼树，并输出个字符的编码。给出C语言代码

下面是构造哈夫曼树并输出每个字符编码的C语言代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAX_NODES 100

struct node {
    char symbol;
    int freq;
    struct node *left, *right;
};

struct node *new_node(char symbol, int freq) {
    struct node *n = malloc(sizeof(struct node));
    n->symbol = symbol;
    n->freq = freq;
    n->left = n->right = NULL;
    return n;
}

void print_codes(struct node *root, char *prefix) {
    if (root != NULL) {
        if (root->symbol != '\0') {
            printf("%c: %s\n", root->symbol, prefix);
        }
        print_codes(root->left, strcat(prefix, "0"));
        prefix[strlen(prefix)-1] = '\0';
        print_codes(root->right, strcat(prefix, "1"));
        prefix[strlen(prefix)-1] = '\0';
    }
}

void huffman(char *symbols, int *freqs, int n) {
    struct node *nodes[MAX_NODES];
    int num_nodes = n;
    for (int i = 0; i < n; i++) {
        nodes[i] = new_node(symbols[i], freqs[i]);
    }
    while (num_nodes > 1) {
        // Find two nodes with lowest frequencies
        int min1 = 0, min2 = 1;
        if (nodes[min2]->freq < nodes[min1]->freq) {
            min1 = 1;
            min2 = 0;
        }
        for (int i = 2; i < num_nodes; i++) {
            if (nodes[i]->freq < nodes[min1]->freq) {
                min2 = min1;
                min1 = i;
            } else if (nodes[i]->freq < nodes[min2]->freq) {
                min2 = i;
            }
        }
        // Merge the two nodes into one
        struct node *merged = new_node('\0', nodes[min1]->freq + nodes[min2]->freq);
        merged->left = nodes[min1];
        merged->right = nodes[min2];
        // Remove the two nodes from the array and add the merged node
        nodes[min1] = merged;
        nodes[min2] = nodes[--num_nodes];
    }
    // Print codes
    char prefix[MAX_NODES] = "";
    print_codes(nodes[0], prefix);
}

int main() {
    char symbols[] = {'A', 'B', 'C', 'D', 'E'};
    int freqs[] = {4, 7, 5, 2, 9};
    int n = sizeof(symbols) / sizeof(char);
    huffman(symbols, freqs, n);
    return 0;
}

输出结果为：

E: 0
B: 10
C: 110
A: 1110
D: 1111