Leetcode-109. Convert Sorted List to Binary Search Tree

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前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

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

这个题目就比较尴尬了，感觉Leetcode是不是抽风了，我对于空输入明明有处理，非说我没处理，然后虽然我效率不高，也误报了一次TLE，结果我看了下论坛的解法，和我差不多。应该是有更快的方法，因为是有序的，每个index的位置是可以算出来的。Your runtime beats 9.60% of java submissions.

public class Solution {    public TreeNode sortedListToBST(ListNode head) {if(null == head) return null;int size = 0;ListNode searchNode = head;while(searchNode != null){ size ++; searchNode = searchNode.next;}if(size == 0) return null;return generateBST(head,size);    }    private TreeNode generateBST(ListNode head,int size){if(size == 0) return null;if(size == 1) return new TreeNode(head.val);if(2 == size){TreeNode root = new TreeNode(head.val);TreeNode right = new TreeNode(head.next.val);root.right = right;return root;}else{int mid = size/2 + 1;ListNode searchNode = head;for(int i = mid; i > 1;i-- ){searchNode = searchNode.next;}TreeNode root = new TreeNode(searchNode.val);TreeNode left = generateBST(head,mid - 1);TreeNode right = generateBST(searchNode.next,size - mid);root.left = left;root.right = right;return root;}    }}

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