C语言实现哈夫曼树求解及其编码输出

目标

给定一组权值，根据权值求其一个哈夫曼树，并通过中序遍历的顺序输出叶子节点的哈夫曼编码。

分析

首先回顾哈夫曼树的求解过程：

在权值中取最小的两个x,y，以这两个权值为叶子节点，生成一个权值为x+y的父亲节点；
从权值组中删除x，y，添加x+y进入权值组；
循环直到权值组仅剩一个元素，该元素为根节点的权值。

一个重要的点是我们希望输入的权值有序，并且时刻保持有序，因此选用链表来存储权重，并每次插入时都按序排序，接下来只要用代码实现上述步骤便可完成哈夫曼树的构建；

那么如何得到叶子节点的哈夫曼编码呢？

考虑使用递归为每个节点都分配一个编码，储存在树节点中。

代码实现

树节点和链表节点的定义与初始化

struct node
{int weight, idx;struct node *parent, *lchild, *rchild;char *code;
};struct linknode
{int info;struct linknode *next;struct node *tree;
};
typedef struct linknode *Plink;
typedef struct node *Ptree;Ptree createNULLtree()
{Ptree head = (Ptree)malloc(sizeof(struct node));if (head == NULL)printf("fail to create tree node!\n");head->parent = NULL;head->lchild = NULL;head->rchild = NULL;head->idx = 0;head->code = (char *)malloc(sizeof(char) * 10);return head;
}Plink createNULLlink()
{Plink head = (Plink)malloc(sizeof(struct linknode));if (head == NULL)printf("fail to create link node!\n");head->next = NULL;head->tree = NULL;return head;
}

树节点中包含指向父亲节点和孩子节点的指针、权值、以及储存着编码的code和编码长度索引idx；

链表节点中包含权值大小info、指向下个节点的next指针以及指向权值对应树节点的tree指针。

一些工具函数

void insertwithorder(Plink head, int x, Ptree tree)
{Plink cur = head->next, last = head;if (tree == NULL)tree = createNULLtree();tree->weight = x;while (cur != NULL){if (x <= cur->info){last->next = createNULLlink();last->next->info = x;last->next->tree = tree;last->next->next = cur;return;}last = cur;cur = cur->next;}last->next = createNULLlink();last->next->info = x;last->next->tree = tree;last->next->next = cur;
}

将数据有序插入链表，小数据在前，插入和删除都比较方便

int is_over(Plink head)
{if (head->next->next == NULL)return 1;return 0;
}

判断是否应该结束程序，当链表中仅剩一个数据时结束

void LDR(Ptree head)
{if (head == NULL)return;LDR(head->lchild);visitleaf(head);LDR(head->rchild);
}

中序遍历

void visitleaf(Ptree head)
{if (head->lchild == NULL && head->rchild == NULL)printf("%d %s\n", head->weight, head->code);
}

只打印叶子节点的信息

生成二叉树

使用两个权值来创建一个二叉树，根节点权值为两权值之和；

实现过程中由于不仅仅需要链表节点的权值信息，所以没有单独使用函数实现这个这个功能；

大致流程如下：

获得两个权值及它们的树节点；
生成一个权值为其之和的树节点；
以新的树节点为父亲节点，连接这三个节点；
将求和之后的权值再按照顺序插入链表中；
使用递归，直到链表中仅剩一个元素；
至此，获得了哈夫曼树的根节点；

void createBiTree(Plink Link)
{if (is_over(Link))return;int nums[2]; //储存两个权值Plink temp = Link->next;Ptree p = createNULLtree(), lch, rch;nums[0] = temp->info;lch = temp->tree;Link->next = temp->next;free(temp);temp = Link->next;nums[1] = temp->info;rch = temp->tree;Link->next = temp->next;free(temp);p->weight = nums[0] + nums[1];p->lchild = lch;p->rchild = rch;lch->weight = nums[0];rch->weight = nums[1];lch->parent = p;rch->parent = p;insertwithorder(Link, p->weight, p);createBiTree(Link);
}

哈夫曼编码

void encode(Ptree head, char info)
{if (head == NULL)return;if (head->parent != NULL){strcpy(head->code, head->parent->code);head->idx = strlen(head->code);}head->code[head->idx++] = info;head->code[head->idx] = '\0';encode(head->lchild, '0');encode(head->rchild, '1');
}

encode函数接收树节点和一个字符变量，左子树接收’0’，右子树接收‘1’，并且每个树节点会copy父亲节点的code信息，如此递归，便完成了编码。

主函数

int main()
{int num, temp;Plink linkhead = createNULLlink();scanf("%d", &num);for (int i = 0; i < num; i++){scanf("%d", &temp);insertwithorder(linkhead, temp, NULL);}createBiTree(linkhead);Ptree treehead = linkhead->next->tree;if (treehead == NULL)printf("wrong!");encode(treehead, '\0');LDR(treehead);return 0;
}

至此便完成了题目要求。

全部代码

#include <stdio.h>
#include <stdlib.h>
#include <string.h>struct node
{int weight, idx;struct node *parent, *lchild, *rchild;char *code;
};struct linknode
{int info;struct linknode *next;struct node *tree;
};
typedef struct linknode *Plink;
typedef struct node *Ptree;Ptree createNULLtree()
{Ptree head = (Ptree)malloc(sizeof(struct node));if (head == NULL)printf("create tree wrongly!\n");head->parent = NULL;head->lchild = NULL;head->rchild = NULL;head->idx = 0;head->code = (char *)malloc(sizeof(char) * 10);return head;
}Plink createNULLlink()
{Plink head = (Plink)malloc(sizeof(struct linknode));if (head == NULL)printf("create link wrongly!\n");head->next = NULL;return head;
}void insertwithorder(Plink head, int x, Ptree tree)
{Plink cur = head->next, last = head;if (tree == NULL)tree = createNULLtree();tree->weight = x;while (cur != NULL){if (x <= cur->info){last->next = createNULLlink();last->next->info = x;last->next->tree = tree;last->next->next = cur;return;}last = cur;cur = cur->next;}last->next = createNULLlink();last->next->info = x;last->next->tree = tree;last->next->next = cur;
}int is_over(Plink head)
{if (head->next->next == NULL)return 1;return 0;
}void createBiTree(Plink Link)
{if (is_over(Link))return;int nums[2];Plink temp = Link->next;Ptree p = createNULLtree(), lch, rch;nums[0] = temp->info;lch = temp->tree;Link->next = temp->next;free(temp);temp = Link->next;nums[1] = temp->info;rch = temp->tree;Link->next = temp->next;free(temp);p->weight = nums[0] + nums[1];p->lchild = lch;p->rchild = rch;lch->weight = nums[0];rch->weight = nums[1];lch->parent = p;rch->parent = p;insertwithorder(Link, p->weight, p);createBiTree(Link);
}void encode(Ptree head, char info)
{if (head == NULL)return;if (head->parent != NULL){strcpy(head->code, head->parent->code);head->idx = strlen(head->code);}head->code[head->idx++] = info;head->code[head->idx] = '\0';encode(head->lchild, '0');encode(head->rchild, '1');
}void visitleaf(Ptree head)
{if (head->lchild == NULL && head->rchild == NULL)printf("%d %s\n", head->weight, head->code);
}void LDR(Ptree head)
{if (head == NULL)return;LDR(head->lchild);visitleaf(head);LDR(head->rchild);
}int main()
{int num, temp;Plink linkhead = createNULLlink();scanf("%d", &num);for (int i = 0; i < num; i++){scanf("%d", &temp);insertwithorder(linkhead, temp, NULL);}createBiTree(linkhead);Ptree treehead = linkhead->next->tree;if (treehead == NULL)printf("wrong!");encode(treehead, '\0');LDR(treehead);return 0;
}