LeetCode Convert Sorted List to Binary Search Tree

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原题链接在这里：https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/

找到中点当BST的root，如此递归调用，知道只有一个点时返回由当前点生成的TreeNode，以此设为终止条件。

Note： 1. 此处midList找到的是中点的前一个点，然后拆成三段，中点前一段，中点单独一个，中点后一段。

2. 递归调用时root.left 就是等于用head为argument的sortListToBST函数，因为如果进了右段，右端的头就是新的head，而不再是原有list的head.

本题与Convert Sorted Array to Binary Search Tree相似。

AC Java:

/** * 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;        }        if(head.next == null){            TreeNode tn = new TreeNode(head.val);            return tn;        }        ListNode middleLeft = midList(head);        ListNode middle = middleLeft.next;        ListNode middleRight = middle.next;                TreeNode root = new TreeNode(middle.val);        middleLeft.next = null;        middle.next = null;        root.left = sortedListToBST(head);        root.right = sortedListToBST(middleRight);        return root;    }    private ListNode midList(ListNode head){        ListNode walker = head;        ListNode runner = head;        while(runner.next != null && runner.next.next != null && runner.next.next.next != null){            runner = runner.next.next;            walker = walker.next;        }        return walker;    }}

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