algorithm - 查找范围内包含的 bst 的最大子树的大小

Question

这是最近的一个面试问题。该问题要求找到包含在 [x, y] 范围内的 BST 的最大子树的大小。是x < y . BST 是递归定义的，其中每个节点都有一个整数值、一个左子节点和一个右子节点。我只能知道树中位于该范围内的节点总数，但无法找到最大的子树。这是我在 python 中使用的代码:

def solution(x, y, T):
    if T is None:
        return 0
    size = 0
    if x < T.val:
        size += solution(x, y, T.left)
    if x <= T.val and y >= T.val:
        size += 1
    # The following if statement was my attempt at resetting the count
    # whenever we find a node outside the range, but it doesn't work
    if x > T.val or y < T.val:
        size = 0
    if B > T.x:
        size += solution(A, B, T.right)
    return size

解决方案应该是O(N)其中 N是树中的节点数。

Answer 1

我们可以递归地解决这个问题。我们需要知道每个子树的左右边界(即最小和最大的元素)。如果它位于 [x, y] 范围内，我们可以用当前子树的总大小更新答案。这是一些代码(solution 函数返回一个元组，在答案之上有一些额外的信息。如果只是想让它返回范围内最大子树的大小，你可以将它包裹起来并使用它作为辅助函数)。

def min_with_none(a, b):
    """
    Returns the minimum of two elements.   
    If one them is None, the other is returned.
    """
    if a is None:
        return b
    if b is None
        return a
    return min(a, b)


def max_with_none(a, b):
    """
    Returns the maximum of two elements.   
    If one them is None, the other is returned.
    """
    if a is None:
        return b
    if b is None:
        return a
    return max(a, b)


def solution(x, y, T):
    """
    This function returns a tuple 
    (max size of subtree in [x, y] range, total size of the subtree, min of subtree, max of subtree) 
    """
    if T is None:
        return (0, 0, None, None)

    # Solves the problem for the children recursively
    left_ans, left_size, left_min, _ = solution(x, y, T.left)
    right_ans, right_size, _, right_max = solution(x, y, T.right)

    # The size of this subtree
    cur_size = 1 + left_size + right_size

    # The left border of the subtree is T.val or the smallest element in the
    # left subtree (if it's not empty)
    cur_min = min_with_none(T.val, left_min)

    # The right border of the subtree is T.val or the largest element in the 
    # right subtree (if it's not empty)
    cur_max = max_with_none(T.val, right_max)

    # The answer is the maximum of answer for the left and for the right 
    # subtree
    cur_ans = max(left_ans, right_ans)
    # If the current subtree is within the [x, y] range, it becomes the new answer,
    # as any subtree of this subtree is smaller than itself
    if x <= cur_min and cur_max <= y:
        cur_ans = cur_size 

    return (cur_size, cur_ans, cur_min, cur_max)

这个解决方案显然以线性时间运行，因为它只访问每个节点一次，并且每个节点执行恒定数量的操作。