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109.有序链表转换二叉搜素树.py
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109.有序链表转换二叉搜素树.py
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# 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 findMiddle(self,head):
prevPtr = None
slowPtr = head
fastPtr = head
# 迭代直到快指针到达了链表的结尾
while fastPtr and fastPtr.next:
prevPtr = slowPtr
slowPtr = slowPtr.next
fastPtr = fastPtr.next.next
# 将链表从慢指针前面断开
if prevPtr:
prevPtr.next = None
return slowPtr
def sortedListToBST(self, head: ListNode) -> TreeNode:
# 边界情况
if not head:
return None
mid = self.findMiddle(head)
# 中间的节点变为BST的根节点
node = TreeNode(mid.val)
# 如果只有一个链表元素
if head == mid:
return node
# 节点的左子树为递归调用之前的头结点
node.left = self.sortedListToBST(head)
# 结点的右子树为递归调用mid.next
node.right = self.sortedListToBST(mid.next)
return node