106. Convert Sorted List to Binary Search Tree
给出一个所有元素以升序排序的单链表，将它转换成一棵高度平衡的二分查找树
样例
2
1->2->3 => / \
1 3实现思路：
如样例给出的一样，平衡二叉树要左右对齐，所以我们可以以中心结点为根结点，对左右部分进行递归建树。
"""
Definition of ListNode
class ListNode(object):
def __init__(self, val, next=None):
self.val = val
self.next = next
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
"""
class Solution:
"""
@param: head: The first node of linked list.
@return: a tree node
"""
def sortedListToBST(self, head):
# write your code here
if head is None:
return None
return self.helper(head)
def helper(self,head):
if head.next is None:
return TreeNode(head.val)
if head.next.next is None:
root = TreeNode(head.val)
root.left = None
root.right = TreeNode(head.next.val)
return root
#获取中心指针和中心指针的前一个指针
mid_pre, mid = self.findPreMid(head)
root = TreeNode(mid.val)
#递归中心指针右半部分
root.right = self.helper(mid.next)
# 移除中心指针和右半部分，就获得了左半部分
mid_pre.next = None
#递归中心指针左半部分
root.left = self.helper(head)
return root
def findPreMid(self, head):
slow, fast = head, head.next
slow_pre = slow
#利用快慢指针,获取中心指针
while fast != None and fast.next != None:
slow_pre = slow
fast = fast.next.next
slow = slow.next
return slow_pre, slow精简版本
#精简版本
class Solution:
"""
@param: head: The first node of linked list.
@return: a tree node
"""
def sortedListToBST(self, head):
# write your code here
if head is None:
return None
if head.next is None:
return TreeNode(head.val)
slow, fast = head, head.next
while fast.next and fast.next.next:
slow = slow.next
fast = fast.next.next
mid = slow.next
slow.next = None
root = TreeNode(mid.val)
root.left = self.sortedListToBST(head)
root.right = self.sortedListToBST(mid.next)
return root