Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
给出一个所有元素以升序排序的单链表,将它转换成一棵高度平衡的二分查找树
/** * 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; } * } */ public class Solution { private ListNode current; public int getListLength(ListNode head) { if (head == null) { return 0; } int size = 0; ListNode temp = head; while (temp != null) { size++; temp = temp.next; } return size; } public TreeNode getCurrentTreeNodeHelper(int size) { if (size == 0) { return null; } TreeNode left = getCurrentTreeNodeHelper(size / 2); TreeNode root = new TreeNode(current.val); current = current.next; TreeNode right = getCurrentTreeNodeHelper(size - 1 - size / 2); root.left = left; root.right = right; return root; } public TreeNode sortedListToBST(ListNode head) { current = head; int size = getListLength(head); TreeNode dummy = getCurrentTreeNodeHelper(size); return dummy; } }
时间: 2024-10-10 01:51:14