递归与二分法

import os
def func(lujing, n): # \"d:/a/\"
    lst = os.listdir(lujing) # 打开文件夹. 列出该文件夹内的所有文件名
    for el in lst: # el是文件的名字.  b, c
        path = os.path.join(lujing, el)  #  还原文件路径\"d:/a/b\"
        if os.path.isdir(path): # 判断路径是否是文件夹
            print(\"...\" * n,el) # 显示文件夹的名字
            func(path, n + 1)  # 在来一次 
        else:
            print(\"\\t\" * n,el) # 显示文件
func(\"d:/a\", 0)

lst = [1,3,5,7,12,36,68,79,132,345,567]
n = int(input(\"请输入一个数\"))
left = 0
right = len(lst) - 1
while left <= right:
    mid = (left + right)//2
    if n > lst[mid]:
        left = mid + 1
    elif n < lst[mid]:
        right = mid - 1
    else:
        print(\"存在\")
        break
else:
    print(\"不存在\")

def func(n, lst):
    left = 0
    right = len(lst) - 1
    if lst != []:
        mid = (left + right)//2
        if n > lst[mid]:
            func(n, lst[mid+1:]) # 改变列表
        elif n < lst[mid]:
            func(n, lst[:mid])
        else:
            print(\"找到了\")
            return
    else:
        print(\"没找到\")
        return

n = int(input(\"请输入你要查找的数:\"))
func(n, [1,3,5,7,12,36,54,68,79,85,92,106]) 

def func(n, lst, left, right): 
    if left <= right:
        mid = (left + right) // 2
        if n > lst[mid]:
            left = mid + 1
            return func(n, lst, left, right)
        elif n < lst[mid]:
            right = mid - 1
            return func(n, lst, left, right) # 递归如果有返回值. 所有调用递归的地方必须写return
        else:
            print(\"找到了\")
            return mid  # 返回上一个调用函数的值
    else:
        print(\"找不到\")
        return False
n = int(input(\"请输入你要查找的数:\"))
lst = lst = [1,3,5,7,12,36,68,79,125,343,485,875,1948]
ret = func(n, lst, 0, len(lst)-1) 
print(ret)