递归返回关键字为key的结点，若未找到返回null（C语言实现）

思路：把h往后调用，当h为0时，返回，返回的值给p，判断h->date==key，若等于把让p指向该位置，返回p。

递归方程：
h0 return NULL;
h!=0 p=FindKeyNode(h->next,key);
if(h->datekey)
p=h;
return p;
运行环境：VS2017
代码实现

#include<stdio.h>
#include<stdlib.h>
typedef struct node
{
	int date;
	struct node*next;
}ElemSN;
//创建链表
ElemSN*GreatLink(int Date[], int n)
{
	int i;
	ElemSN*p, *t = 0, *h = 0;
	for (i = 0;i < n;i++)
	{
		p = (ElemSN*)malloc(sizeof(ElemSN));
		p->date = Date[i];
		p->next = NULL;
		if (!h)
			h = t = p;
		else
			t = t->next = p;
	}
	return h;
}
//输出链表
PrintLink(ElemSN*h)
{
	ElemSN*p;
	for (p = h;p;p = p->next)
		printf("%3d", p->date);
	printf("\n");
}

ElemSN*FindKeyNode(ElemSN*h, int key)
{
	ElemSN*p = NULL;
	if (h)
	{
		p = FindKeyNode(h->next, key);
		if (h->date == key)
			p = h;
	}
	return p;
}

int main(void)
{
	int a[8] = { 3,2,5,8,4,7,6,9 };
	ElemSN*head, *p;
	head = GreatLink(a, 8);
	PrintLink(head);
	p = FindKeyNode(head, 7);
	printf("%3d\n", p->date);
	system("pause");
}

运行结果

  3  2  5  8  4  7  6  9
  7
请按任意键继续. . .