【LeetCode】 109. Convert Sorted List to Binary Search Tree 有序链表转换二叉搜索树（Medium）（JAVA）

【LeetCode】 109. Convert Sorted List to Binary Search Tree 有序链表转换二叉搜索树（Medium）（JAVA）

题目地址: https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/

题目描述：

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

题目大意

给定一个单链表，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的绝对值不超过 1。

解题方法

1、和上一题类似，只是不像数组，可以知道长度和某个位置的元素 【LeetCode】 108. Convert Sorted Array to Binary Search Tree 将有序数组转换为二叉搜索树（Easy）（JAVA）
2、采用快慢指针，找出中间节点
3、采用递归

/**
 * 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 slow = head;
        ListNode fast = head.next.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        TreeNode root = new TreeNode(slow.next.val);
        ListNode right = slow.next.next;
        slow.next = null;
        root.left = sortedListToBST(head);
        root.right = sortedListToBST(right);
        return root;
    }
}

执行用时 : 0 ms, 在所有 Java 提交中击败了 100.00% 的用户
内存消耗 : 40.5 MB, 在所有 Java 提交中击败了 13.33% 的用户