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.

由于已经有序，所以利用二分法创建这棵平衡二叉树，与Sort List类似

需要注意的就是一条链的后半条链的索引范围是[0，high - mid - 1]，原来一直纠结在high - mid;

其次，就是注意后半条链的链头。

/**
 * 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) {
        int len = 0;
        ListNode p = head;
        while(p != null){
            len++;
            p = p.next;
        }
        return create(head,0,len - 1);
    }
    public TreeNode create(ListNode head, int low , int high){
        if(low > high ){
            return null;
        }
        int mid = (low + high) >> 1;
        int count = 0;
        ListNode p = head;
        while(count < mid){
            p = p.next;
            count++;
        }
        TreeNode t = new TreeNode(p.val);
        t.left = create(head,0,mid-1);
        t.right = create(p.next,0,high - mid - 1);
        return t;
    }
}

Runtime: 281 ms