# 0109. Convert Sorted List to Binary Search Tree

<https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree>

## Description

Given the `head` of 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 1:**

![](https://assets.leetcode.com/uploads/2020/08/17/linked.jpg)

```
**Input:** head = [-10,-3,0,5,9]
**Output:** [0,-3,9,-10,null,5]
**Explanation:** One possible answer is [0,-3,9,-10,null,5], which represents the shown height balanced BST.
```

**Example 2:**

```
**Input:** head = []
**Output:** []
```

**Example 3:**

```
**Input:** head = [0]
**Output:** [0]
```

**Example 4:**

```
**Input:** head = [1,3]
**Output:** [3,1]
```

**Constraints:**

* The number of nodes in `head` is in the range `[0, 2 * 104]`.
* `-105 <= Node.val <= 105`

## ac

```java
/**
 * 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) {
        // corner case, exit
        if (head == null) return null; // len == 0
        if (head.next == null) return new TreeNode(head.val); // len == 1

        // find mid
        ListNode slow = head, fast = head;
        ListNode prev = new ListNode(0);
        prev.next = head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            prev = prev.next;
        }
        // seperate
        ListNode h2 = slow.next;
        slow.next = null;
        prev.next = null;
        // divide & conquer
        TreeNode root = new TreeNode(slow.val);
        root.left = sortedListToBST(head);
        root.right = sortedListToBST(h2);

        return root;
    }
}
```