convert-sorted-list-to-binary-search-tree

最新推荐文章于 2022-05-08 09:14:34 发布

原创最新推荐文章于 2022-05-08 09:14:34 发布 · 264 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#链表

算法专栏收录该内容

10 篇文章

订阅专栏

本文介绍了一种将升序排列的单链表转换为高度平衡的二叉搜索树的方法。通过递归地选择中间节点作为根节点来确保树的平衡性。

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

class TreeNode(object):
   def __init__(self, val, left=None, right=None):
      self.data = val
      self.left = left
      self.right = right


class Solution(object):
   def toBST(self, head, tail):
      if head != tail or head.next != tail:
         return head
      slow = head
      fast = head
      while fast is not None and fast.next is not None:
         slow = slow.next
         fast = fast.next.next
      root = TreeNode(head.data)
      root.left = self.toBST(head, slow)
      root.right = self.toBST(slow.next, fast)
      return root

   def sortedListToBST(self, head):
      return self.toBST(head, None)