LeetCode-109、 有序链表转换二叉搜索树-中等
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
代码:
# 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 tree(cur):
if not cur.next:
return TreeNode(cur.val)
pre = ListNode(None)
pre.next = cur
slow, fast = cur, cur
slow_pre = pre
while fast and fast.next:
slow_pre = slow_pre.next
slow = slow.next
fast = fast.next.next
slow_pre.next = None
node = TreeNode(slow.val)
if cur:
node.left = tree(cur)
if slow.next:
node.right = tree(slow.next)
return node
if head is None:
return head
else:
return tree(head)