Convert Sorted List to Binary Search Tree 把有序链表转为BST@LeetCode

万能的递归法，注意到有一点比较巧的是，由于单链表都是单向的。所以这里我的查找区间设置为左闭右开！这样就避免了去找mid节点的前一个节点

package Level4;import Utility.ListNode;import Utility.TreeNode;/** * Convert Sorted List to Binary Search Tree *  *  Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. * */public class S109 {public static void main(String[] args) {ListNode head = new ListNode(1);ListNode n2 = new ListNode(3);head.next = n2;TreeNode root = sortedListToBST(head);}public static TreeNode sortedListToBST(ListNode head) {        return rec(head, null);    }// 在区间[start, end)里递归，后面的end是包括在内的，这样可以避免要多用一个指针来记录mid前的节点public static TreeNode rec(ListNode start, ListNode end){if(start == end){return null;}// 一次遍历找到中点的方法：快慢指针ListNode mid = start;// 该指针最终会指向中点ListNode probe = start;// 探针最终会到达endwhile(probe!=end && probe.next!=end){// 探针完成搜索，注意停止条件是和end比较而不是和null比！mid = mid.next;probe = probe.next.next;}TreeNode root = new TreeNode(mid.val);root.left = rec(start, mid);root.right = rec(mid.next, end);return root;}}

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; next = null; } * } *//** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public TreeNode sortedListToBST(ListNode head) {        return rec(head, null);    }        public TreeNode rec(ListNode start, ListNode end) {        if(start == end) {            return null;        }        ListNode p = start, q = start;        while(q != end && q.next != end) {            p = p.next;            q = q.next.next;        }                TreeNode root = new TreeNode(p.val);        root.left = rec(start, p);        root.right = rec(p.next, end);                return root;    }    }