在一次函数执行过程中只会执行一次return.
int f(int k)
{
if(k==1)
return 1;
else
return k*f(k-1); 可以解释一下这里的两个return的返回值是怎样返回的吗？
}
每次递归调用时判断递归是否到了出口(k==1).到了则执行return 1.否则该函数下执行返回k*(f(k-1))
这里它还会继续递归调用f(k-1)直到f(1)到出口.

int SUM(int n)
{int sum;
if(n==1)
sum=1;
else
sum=n+SUM(n-1);
return sum;
}

这个比较简单!!

最典型的例子应该是汉诺塔
而像fibonacci数列有更好的方式实现
我觉得如果能完全理解汉诺塔的递归程序,那么可以说已经对递归有一定了解了

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

extern int cnt; /* count of the number of moves */

int get_n_from_user(void);
void move(int n, char a, char b, char c);

#include "hanoi.h"

int cnt = 0; /* count of the number of moves */

int main(void)
{
int n;

n = get_n_from_user();
assert(n > 0);
/*
// Move n disks from tower A to tower C,
// using tower B as an intermediate tower.
*/
move(n, 'A', 'B', 'C'); /* recursive fct */
return 0;
}

#include "hanoi.h"

void move(int n, char a, char b, char c)
{
if (n == 1) {
++cnt;
printf("%5d: %s%d%s%c%s%c.\n", cnt,
"Move disk ", 1, "from tower ", a, " to tower ", c);
}
else {
move(n - 1, a, c, b);
++cnt;
printf("%5d: %s%d%s%c%s%c.\n", cnt,
"Move disk ", n, "from tower ", a, " to tower ", c);
move(n - 1, b, a, c);
}
}

#include "hanoi.h"

int get_n_from_user(void)
{
int n;

printf("%s",
"---\n"
"TOWERS OF HANOI:"
"\n"
"There are three towers: A, B, and C.\n"
"\n"
"The disks on tower A must be moved to tower C. Only one\n"
"disk can be moved at a time, and the order on each tower\n"
"must be preserved at a time, and the order on each tower\n"
"or C can be used for intermediate placement of a disk.\n"
"\n"
"The problem starts with n disk on Tower A.\n"
"\n"
"Input n: ");
if (scanf("%d", &n) != 1 || n < 1) {
printf("\nERROR: Positive integer not found - bye!\n\n");
exit(1);
}
printf("\n");
return n;
}

这里面是不是包括了二叉树的左遍历的知识