极客算法

LeetCode-109.有序链表转换二叉搜索树(Convert Sorted List to Binary Search Tree)


109. 有序链表转换二叉搜索树

给定一个单链表,其中的元素按升序排序,将其转换为高度平衡的二叉搜索树。

本题中,一个高度平衡二叉树是指一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1。

图片说明

示例:

给定的有序链表: [-10, -3, 0, 5, 9],

一个可能的答案是:[0, -3, 9, -10, null, 5], 它可以表示下面这个高度平衡二叉搜索树:

      0
     / \
   -3   9
   /   /
 -10  5

来源:力扣(LeetCode)

链接:https://leetcode-cn.com/problems/convert-sorted-list-to-binary-search-tree/

Link:https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/

递归 + 双指针

O(N)

BST的中序遍历的结果是一个不下降的序列。找到中间节点,左边小于的都是左子树,右边大于的都是右子树

这道题,和上一道题是一样的,只是对应的数据结构是链表。数组很容易找到中间的那个元素,链表如何找到中间节点,就成了最直接的问题

链表的中间节点,使用快慢指针。但要记住,一定要把指针断开

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def sortedListToBST(self, head: ListNode) -> TreeNode:
        left, cur, right = self.splitLinkedList(head)
        
        root = None
        
        if cur is not None:
            root = TreeNode(cur.val)

            root.left = self.sortedListToBST(left)
            root.right = self.sortedListToBST(right)
                
        return root

    def splitLinkedList(self, head: ListNode) -> Tuple[ListNode, ListNode, ListNode]:
    
        if head is None:
            return (None, None, None)
    
        sentryHead = ListNode(0)
        sentryHead.next = head
        
        fast = sentryHead
        slow = sentryHead

        while fast.next is not None and fast.next.next is not None:
            fast = fast.next.next
            slow = slow.next
            
        cur = slow.next
        right = slow.next.next
        slow.next = None # 断开左右两部分
        
        return (sentryHead.next, cur, right)

–End–


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