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.

和Convert Sorted Array to Binary Search Tree的区别是array可以用二分法，而list没法直接算mid。这时可用双指针，一个走快一个走慢的方法，找mid，然后再递归。

Source

/** * 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) {        if(head == null) return null;                ListNode p = head;        ListNode q = head;        ListNode pre = null;                while(q.next!= null && q.next.next != null){  //注意判断条件         pre = p;        p = p.next;           q = q.next.next;        }        if(pre != null) pre.next = null;  //注意pre==null的情况                TreeNode root = new TreeNode(p.val);        if(pre != null) root.left = sortedListToBST(head);  //注意pre==null的情况        root.right = sortedListToBST(p.next);        return root;    }}

Test

    public static void main(String[] args){    ListNode head = new ListNode(1);    head.next = new ListNode(2);    head.next.next = new ListNode(3);    head.next.next.next = new ListNode(4);    head.next.next.next.next = new ListNode(5);    TreeNode a = new Solution().sortedListToBST(head);            dfs(a);    }    public static void dfs(TreeNode a){    if(a == null) return;    System.out.println(a.val);        dfs(a.left);    dfs(a.right);    }