題目如下：

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of?every?node never differ by more than 1.

Example:

Given the sorted linked list: [-10,-3,0,5,9],One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:0/ \-3 9/ /-10 5

解題思路：題目沒有明確要求不允許使用額外的內存，所以最簡單的方法是把linked list中每個元素的值存入list，然后套用【leetcode】108. Convert Sorted Array to Binary Search Tree?的解法。

代碼如下：

# Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = Noneclass Solution(object):def recursive(self,node,nums):mid = len(nums)/2left_num = nums[:mid]if len(left_num) > 0:node.left = TreeNode(left_num[len(left_num)/2])self.recursive(node.left,left_num)right_num = nums[mid+1:]if len(right_num) > 0:node.right = TreeNode(right_num[len(right_num)/2])self.recursive(node.right,right_num)def sortedListToBST(self, head):""":type head: ListNode:rtype: TreeNode"""nums = []while head != None:nums.append(head.val)head = head.nextif len(nums) == 0:return Noneroot = TreeNode(nums[len(nums)/2])self.recursive(root,nums)return root

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轉載于:https://www.cnblogs.com/seyjs/p/10625312.html

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