algorithm – 查找范围中包含的bst的最大子树的大小

2019年7月28日 226次阅读

这是最近一次采访问题.要求找到包含在[x,y]范围内的BST的最大子树的大小的问题是x <1.年. BST是递归定义的,其中每个节点具有整数值,左子节点和右子节点.我只能在树中的节点总数位于该范围内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是树中节点的数量.

最佳答案 我们可以递归地解决问题.我们需要知道每个子树的左边界和右边界(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.下面是一些代码(解决方案函数在答案之上返回一个带有一些额外信息的元组.如果只是希望它返回范围内最大子树的大小,你可以将它包装起来并将其用作辅助函数).

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)

该解决方案明确地以线性时间运行,因为它仅访问每个节点一次并且每个节点执行恒定数量的操作.

点赞