哈夫曼树代码实现

写了我1个小时wok

哈夫曼树的构建先要找两个最小的两个数

int* minxx(struct HFtree* HF)
{
	int min = 99999;
	int minindex=-1;
	int mintwoindex = -1;
	int* res = (int*)malloc(sizeof(int) * 2);
	for (int i = 0; i < HF->len; i++)
	{
		if (HF->data[i].dad == 0)
		{
			if (HF->data[i].val < min)
			{
				min = HF->data[i].val;
				minindex = i;
			}
		}
		res[0] = minindex;
	}
	min = 99999;
	for (int i = 0; i < HF->len; i++)
	{
		if (HF->data[i].dad == 0&&i!=minindex)
		{
			if (HF->data[i].val < min)
			{
				min = HF->data[i].val;
				mintwoindex = i;
			}
		}
		res[1] = mintwoindex;
	}
	return res;
}

然后就是这个树的初始化，我们先要让他的父节点都是0，左右子节点都是-1

struct HFtree* start(int len,int* a)
{
	struct HFtree* HF = (struct HFtree*)malloc(sizeof(struct HFtree));
	HF->data = (Node*)malloc(sizeof(Node) * (2 * len - 1));
	HF->len = len;
	for (int i = 0; i < len; i++)
	{
		HF->data[i].val = a[i];
		HF->data[i].left = -1;
		HF->data[i].right = -1;
		HF->data[i].dad = 0;
	}
	return HF;
}

然后就是他的树的建立

void creat(struct HFtree* HF)
{
	int* res;
	int len = HF->len * 2 - 1;
	for (int i = HF->len; i < len; i++)
	{
		res = minxx(HF);
		HF->data[i].dad = 0;
		HF->data[i].val = HF->data[res[0]].val + HF->data[res[1]].val;//父亲的价值为儿子的价值加一起
		HF->data[i].left = res[0];//左儿子为下标
		HF->data[i].right = res[1];
		HF->data[res[0]].dad = i;//左右儿子的父亲为i
		HF->data[res[1]].dad = i;
		HF->len++;//创建出一个就加1
	}
}

先序遍历，如果有左右儿子就一直找

void xianxu(struct HFtree* HF,int index)//先序遍历哈夫曼树
{
	if (index != -1)
	{
		printf("%d ", HF->data[index].val);
		xianxu(HF, HF->data[index].left);
		xianxu(HF, HF->data[index].right);
	}
}

总的代码就是

#include 
#include 
typedef struct node
{
	int val;//该点的价值
	int left, right;//左右节点
	int dad;//父亲节点
}Node;
struct HFtree
{
	Node* data;//数组
	int len;//数组长度
};
struct HFtree* start(int len,int* a)
{
	struct HFtree* HF = (struct HFtree*)malloc(sizeof(struct HFtree));
	HF->data = (Node*)malloc(sizeof(Node) * (2 * len - 1));
	HF->len = len;
	for (int i = 0; i < len; i++)
	{
		HF->data[i].val = a[i];
		HF->data[i].left = -1;
		HF->data[i].right = -1;
		HF->data[i].dad = 0;
	}
	return HF;
}
int* minxx(struct HFtree* HF)
{
	int min = 99999;
	int minindex=-1;
	int mintwoindex = -1;
	int* res = (int*)malloc(sizeof(int) * 2);
	for (int i = 0; i < HF->len; i++)
	{
		if (HF->data[i].dad == 0)
		{
			if (HF->data[i].val < min)
			{
				min = HF->data[i].val;
				minindex = i;
			}
		}
		res[0] = minindex;
	}
	min = 99999;
	for (int i = 0; i < HF->len; i++)
	{
		if (HF->data[i].dad == 0&&i!=minindex)
		{
			if (HF->data[i].val < min)
			{
				min = HF->data[i].val;
				mintwoindex = i;
			}
		}
		res[1] = mintwoindex;
	}
	return res;
}
void creat(struct HFtree* HF)
{
	int* res;
	int len = HF->len * 2 - 1;
	for (int i = HF->len; i < len; i++)
	{
		res = minxx(HF);
		HF->data[i].dad = 0;
		HF->data[i].val = HF->data[res[0]].val + HF->data[res[1]].val;//父亲的价值为儿子的价值加一起
		HF->data[i].left = res[0];//左儿子为下标
		HF->data[i].right = res[1];
		HF->data[res[0]].dad = i;//左右儿子的父亲为i
		HF->data[res[1]].dad = i;
		HF->len++;//创建出一个就加1
	}
}
void xianxu(struct HFtree* HF,int index)//先序遍历哈夫曼树
{
	if (index != -1)
	{
		printf("%d ", HF->data[index].val);
		xianxu(HF, HF->data[index].left);
		xianxu(HF, HF->data[index].right);
	}
}
int main()
{
	int a[4] = { 1,2,3,4 };
	struct HFtree* HF;
	HF = start(4, a);
	creat(HF);//建树也没错
	for (int i = 0; i < 2 * 4 - 1; i++)
	{
		printf("%d ", HF->data[i].val);
		printf("%d ", HF->data[i].left);
		printf("%d ", HF->data[i].right);
		printf("%d ", HF->data[i].dad);
		printf("\n");
	}
	//printf("%d ", HF->data[4 * 2 - 1].left);
	xianxu(HF, HF->len - 1);//最后的一个数为根
	//int* res = minxx(HF);
	//printf("%d %d", res[0], res[1]);
}

还有个哈夫曼树的编码没有搞出来，和二叉树的遍历感觉差不多，只不过向左就是输出0，向右就是输出1，明天在搞

下班下班