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109. Convert Sorted List to Binary Search Tree
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
思路：递归做。不断取中间数，建root。把链表里面的数字都读出来存在list上面进行操作。要多想想这道题是什么样的类型，要用什么方法做比较适合，方便。
/** * 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) { List<Integer> res=new ArrayList<Integer>(); while(head!=null) { res.add(head.val); head=head.next; } return helper(res,0,res.size()-1); } public TreeNode helper(List<Integer> list,int low,int high) { if(low>high) { return null; } int mid=low+(high-low)/2; TreeNode root=new TreeNode(list.get(mid)); root.left=helper(list,low,mid-1); root.right=helper(list,mid+1,high); return root; }}
109. Convert Sorted List to Binary Search Tree
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