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109. Convert Sorted List to Binary Search Tree

时间:2016-10-09 07:20:28      阅读:214      评论:0      收藏:0      [点我收藏+]

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

原文:http://www.cnblogs.com/Machelsky/p/5940761.html

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