[LeetCode] 109. Convert Sorted List to Binary Search Tree

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:

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].
-10^5 <= Node.val <= 10^5

有序链表转换二叉搜索树。

给定一个单链表的头节点 head ，其中的元素按升序排序，将其转换为高度平衡的二叉搜索树。

本题中，一个高度平衡二叉树是指一个二叉树每个节点的左右两个子树的高度差不超过 1。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/convert-sorted-list-to-binary-search-tree
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题目即是题意，可以跟108题一起做。108是从有序数组转化成BST，109是从有序链表转化成BST。区别在于108可以通过找中点的办法快速找到根节点，但是109只能通过快慢指针的办法找到根节点，思路都是递归。

时间O(n)

空间O(n)

Java实现

 1 class Solution {
 2     public TreeNode sortedListToBST(ListNode head) {
 3         if (head == null)
 4             return null;
 5         return helper(head, null);
 6     }
 7 
 8     public TreeNode helper(ListNode head, ListNode tail) {
 9         if (head == tail)
10             return null;
11         ListNode slow = head;
12         ListNode fast = head;
13         while (fast != tail && fast.next != tail) {
14             fast = fast.next.next;
15             slow = slow.next;
16         }
17         TreeNode root = new TreeNode(slow.val);
18         root.left = helper(head, slow);
19         root.right = helper(slow.next, tail);
20         return root;
21     }
22 }

JavaScript实现

 1 /**
 2  * @param {ListNode} head
 3  * @return {TreeNode}
 4  */
 5 var sortedListToBST = function (head) {
 6     if (head === null) return null;
 7     return helper(head, null);
 8 };
 9 
10 var helper = function (head, tail) {
11     if (head === tail) return null;
12     let slow = head;
13     let fast = head;
14     while (fast !== tail && fast.next !== tail) {
15         fast = fast.next.next;
16         slow = slow.next;
17     }
18     let root = new TreeNode(slow.val);
19     root.left = helper(head, slow);
20     root.right = helper(slow.next, tail);
21     return root;
22 }

CNoodle

Let the code speak for you.

[LeetCode] 109. Convert Sorted List to Binary Search Tree

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