Convert Sorted List to Binary Search Tree

lc109 Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Given the sorted linked list: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5

lc 109 思路找中点，除了middle.next 设为null还需要注意moddle的前一个node的next也要设为null

 /**
 * 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; }
 * }
 */
class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        if (head == null) return null;
        if (head.next == null) return new TreeNode(head.val);
        ListNode middle = findMiddle(head);
        TreeNode right = sortedListToBST(middle.next);
        middle.next = null;
        TreeNode root = new TreeNode(middle.val);
        TreeNode left = null;
        if (head != middle) {
            left = sortedListToBST(head);
        }
        root.left = left;
        root.right = right;
        return root;
    }

    public ListNode findMiddle(ListNode node) {
        if (node == null) return node;
        ListNode prev = null;
        ListNode slow = node;
        ListNode fast = node.next;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            prev = slow;
            slow = slow.next;
        }
        if (prev != null) {
            prev.next = null;
        }
        return slow;
    } 
}