python – 如何将以下函数转换为尾递归函数？

我正在跟随here,我正在尝试将正常的递归函数转换为尾递归函数.我设法理解斐波纳契和阶乘版本,但这个让我很难过.我理解算法在做什么,以及在转换中让我困惑的else语句.

在其他内部,它试图找到一个更接近你正在寻找的数字,然后放弃并找到它找到的数字小于你建议的数字.

我不知道如何编写使尾部递归的辅助函数.对于斐波那契和阶乘,我最终使用了累加器.有没有类似的东西可以在这里使用？

class BSTNode(object):
    """Binary search tree node."""

    def __init__(self, val, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

    def __repr__(self):
        return '(%s, %r, %r)' % (self.val, self.left, self.right)

def find_val_or_next_smallest(bst, x):
    """
    Get the greatest value <= x in a binary search tree.
    Returns None if no such value can be found.
    """
    if bst is None:
        return None
    elif bst.val == x:
        return x
    elif bst.val > x:
        return find_val_or_next_smallest(bst.left, x)
    else:
        right_best = find_val_or_next_smallest(bst.right, x)
        if right_best is None:
            return bst.val
        return right_best

我知道Python不支持尾递归优化以允许常量堆栈空间但我只是在Python中练习这样做,因为我喜欢语法

解决方法:

而不是做

if right_best is None:
    return bst.val

您可以将目前为止找到的最佳结果作为额外参数传递给递归调用,并使递归调用处理此检查.

def find_val_or_next_smallest(bst, x, best=None):
    """
    Get the greatest value <= x in a binary search tree.
    Returns None if no such value can be found.
    """
    if bst is None:
        return best
    elif bst.val == x:
        return x
    elif bst.val > x:
        return find_val_or_next_smallest(bst.left, x, best)
    else:
        # bst.val is guaranteed to be the best yet, since if we had
        # seen a better value higher up, the recursion would have gone
        # the other way from that node
        return find_val_or_next_smallest(bst.right, x, bst.val)

标签：python,algorithm,recursion,binary-tree,tail-recursion
来源： https://codeday.me/bug/20190829/1764436.html