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。

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:
        def getMid(head):
            pre=None
            slow=fast=head
            while fast and fast.next:
                pre=slow
                slow=slow.next
                fast=fast.next.next
            if pre:
                pre.next=None
            return slow
        
        if not head:
            return None
        mid=getMid(head)
        node=TreeNode(mid.val)
        if head==mid:
            return node
        node.left=self.sortedListToBST(head)
        node.right=self.sortedListToBST(mid.next)
        return node

 

 

# 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:
        nums=[]
        while head:
            nums.append(head.val)
            head=head.next
        def dfs(nums,l,r):
            if l==r:return 
            mid=l+(r-l)//2
            node=TreeNode(nums[mid])
            node.left=dfs(nums,l,mid)
            node.right=dfs(nums,mid+1,r)
            return node
        return dfs(nums,0,len(nums))