原题地址:
https://oj.leetcode.com/problems/convert-sorted-list-to-binary-search-tree/
题目内容:
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
方法:
单纯。如何安排插入顺序,使得一棵二叉排序树接近平衡?
你从中间开始插嘛。这样左子树和右子树之差或者为0,或者为1。
所以,这个问题的本质是找链表的中间节点。
找到中间节点后,递归产生左子树和右子树,在找左边子链表的中间节点和右边子链表的中间节点即可。
当然啦,最后返回中间节点。
全部代码:
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; next = null; } * } */ /** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public TreeNode sortedListToBST(ListNode head) { int len = countLen(head); if (len == 0) return null; return constructTree(head,len); // len is the number of nodes in list. } private TreeNode constructTree(ListNode head,int len) { if (head == null || len <= 0) return null; int mid = len / 2 + 1; TreeNode tmp = new TreeNode(head.val); tmp.left = null; tmp.right = null; if (mid == 0) return tmp; ListNode tar = findMidNode(head,mid); tmp.val = tar.val; tmp.left = constructTree(head,mid - 1); tmp.right = constructTree(tar.next,len - mid); return tmp; } private ListNode findMidNode(ListNode head,int mid) { int count = 1; while (mid != count++) head = head.next; return head; } private int countLen(ListNode head) { int len = 0; while (head != null) { len ++; head = head.next; } return len; } }
时间: 2024-10-05 00:11:49