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

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题目要求：

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

给定一个排好序的链表，转换成二叉搜索树

思路：和108题类似，先遍历链表元素，存储到数组中，把数组中间的元素作为根节点，数组左边的元素是左子树，右边的元素是右子树，递归创建左子树右子树。

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } *//** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public TreeNode sortedListToBST(ListNode head) {                       if(head==null)        {            return null;        }       ArrayList<Integer> arr = new ArrayList<Integer>();        while(head!=null)        {           arr.add(head.val);            head = head.next;        }        return buildnode(arr,0,arr.size()-1);    }    //建立树    public TreeNode buildnode(ArrayList<Integer> arr,int start,int end)    {        if(start<=end)        {            int mid = (start+end)/2;            TreeNode node = new TreeNode(arr.get(mid));            node.left = buildnode(arr,start,mid-1);            node.right=buildnode(arr,mid+1,end);            return node;        }        return null;    }}

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