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Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
/** * 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 { public TreeNode sortedListToBST(ListNode head) { if(head==null) return null; int len=0; ListNode node=head; while(node!=null){ node=node.next; len++; } return getBST(head,0,len-1); } TreeNode getBST(ListNode head,int start,int end){ if(start>end) return null; int mid=(start+end)/2; ListNode temp=head; for(int i=start;i<mid;i++){//求中间节点方法————start到mid temp=temp.next; } TreeNode node=new TreeNode(temp.val); node.left=getBST(head,start,mid-1); node.right=getBST(temp.next,mid+1,end);//注意右侧节点链表的头结点!!!因为和数组求中间节点的方式不一样,链表主要是遍历,求相对位置 return node; } }
[leedcode 109] Convert Sorted List to Binary Search Tree
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原文地址:http://www.cnblogs.com/qiaomu/p/4665969.html