LeetCode 109. 有序链表转换二叉搜索树

Posted 2020-12-17 机器狗mo

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。

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

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def sortedListToBST(self, head: ListNode) -> TreeNode:
        if head is None:
            return None
        if head.next is None:
            return TreeNode(head.val)
        
        pre_slow = ListNode(None)
        pre_slow.next = head
        slow = head
        fast = head 
        while fast is not None and fast.next is not None:
            pre_slow = pre_slow.next 
            slow = slow.next
            fast = fast.next.next

        mid = pre_slow.next 
        pre_slow.next = None
        mid_node = TreeNode(mid.val)
        mid_node.left = self.sortedListToBST(head)
        mid_node.right = self.sortedListToBST(mid.next)
        return mid_node