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边界和右边界(即最小和最大的元素)。如果它位于范围[x,y]中,我们可以用当前子树的总大小来更新答案。下面是一些代码(solution
函数返回一个元组,其中包含一些额外的答案信息,如果只想返回范围内最大子树的大小,可以将它包装起来并用作辅助函数)。
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)
该溶液显然线性时间运行,因为它访问每一个节点仅一次,并执行每节点的操作的常数。