生成一个二叉树代码

def tree(n):
    if n == 0:
        return None
    else:
        return (tree(n//2), n%2)

def main():
    n = int(input())
    root = tree(n)
    print(root)

if __name__ == '__main__':
    main()

A:

You can use recursion to solve this problem. The idea is to keep track of the current node and the