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	d26e811e10
	
	
	
		
			
			* fix the bugs of C code. * Add a header figure. * Improve the definition of tree node height.
		
			
				
	
	
		
			78 lines
		
	
	
		
			1.6 KiB
		
	
	
	
		
			C
		
	
	
	
	
	
			
		
		
	
	
			78 lines
		
	
	
		
			1.6 KiB
		
	
	
	
		
			C
		
	
	
	
	
	
| /**
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|  * File: recursion.c
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|  * Created Time: 2023-09-09
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|  * Author: Gonglja (glj0@outlook.com)
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|  */
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| 
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| #include "../utils/common.h"
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| 
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| /* 递归 */
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| int recur(int n) {
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|     // 终止条件
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|     if (n == 1)
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|         return 1;
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|     // 递:递归调用
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|     int res = recur(n - 1);
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|     // 归:返回结果
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|     return n + res;
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| }
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| 
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| /* 使用迭代模拟递归 */
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| int forLoopRecur(int n) {
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|     int stack[1000]; // 借助一个大数组来模拟栈
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|     int top = -1;    // 栈顶索引
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|     int res = 0;
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|     // 递:递归调用
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|     for (int i = n; i > 0; i--) {
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|         // 通过“入栈操作”模拟“递”
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|         stack[1 + top++] = i;
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|     }
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|     // 归:返回结果
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|     while (top >= 0) {
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|         // 通过“出栈操作”模拟“归”
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|         res += stack[top--];
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|     }
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|     // res = 1+2+3+...+n
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|     return res;
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| }
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| 
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| /* 尾递归 */
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| int tailRecur(int n, int res) {
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|     // 终止条件
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|     if (n == 0)
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|         return res;
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|     // 尾递归调用
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|     return tailRecur(n - 1, res + n);
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| }
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| 
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| /* 斐波那契数列:递归 */
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| int fib(int n) {
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|     // 终止条件 f(1) = 0, f(2) = 1
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|     if (n == 1 || n == 2)
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|         return n - 1;
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|     // 递归调用 f(n) = f(n-1) + f(n-2)
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|     int res = fib(n - 1) + fib(n - 2);
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|     // 返回结果 f(n)
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|     return res;
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| }
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| 
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| /* Driver Code */
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| int main() {
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|     int n = 5;
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|     int res;
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| 
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|     res = recur(n);
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|     printf("\n递归函数的求和结果 res = %d\n", res);
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| 
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|     res = forLoopRecur(n);
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|     printf("\n使用迭代模拟递归求和结果 res = %d\n", res);
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| 
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|     res = tailRecur(n, 0);
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|     printf("\n尾递归函数的求和结果 res = %d\n", res);
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| 
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|     res = fib(n);
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|     printf("\n斐波那契数列的第 %d 项为 %d\n", n, res);
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| 
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|     return 0;
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| }
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