【leetcode】109. Convert Sorted List to Binary Search Tree

题目如下：

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

解题思路：题目没有明确要求不允许使用额外的内存，所以最简单的方法是把linked list中每个元素的值存入list，然后套用【leetcode】108. Convert Sorted Array to Binary Search Tree 的解法。

代码如下：

# 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

转载于:https://www.cnblogs.com/seyjs/p/10625312.html

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