109. 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.
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.
Example:

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

難度：medium

題目：給定一個單鏈表其元素爲升序排列，將其轉換成高度平衡的二叉搜索樹

思路：中序遍歷

Runtime: 1 ms, faster than 99.17% of Java online submissions for Convert Sorted List to Binary Search Tree.
Memory Usage: 41 MB, less than 9.56% of Java online submissions for Convert Sorted List to Binary Search Tree.

/**
 * 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 (null == head) {
            return null;
        }
        ListNode ptr = head;
        int count = 0;
        for (; ptr != null; ptr = ptr.next, count++);
        ListNode[] headList = {head};
        
        return sortedListToBST(headList, 0, count - 1);
    }
    
    public TreeNode sortedListToBST(ListNode[] head, int start, int end) {
        if (start > end) {
            return null;
        }
        int mid = start + (end - start) / 2;
        TreeNode left = sortedListToBST(head, start, mid - 1);
        TreeNode root = new TreeNode(head[0].val);
        root.left = left;
        head[0] = head[0].next;
        
        root.right = sortedListToBST(head, mid + 1, end);

        return root;
    }
}