转换有序链表成二叉搜索树

来源：互联网发布：mac 查找程序安装路径编辑：程序博客网时间：2024/05/17 06:09

题目原型：

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

基本思路：

先将链表中的数存入数组，转化成有序数组转换为二叉搜索树，对于一个有序数组先找到中间值作为根，左边的为左子树，右边的为右子树，以此递归即可。

    public TreeNode sortedListToBST(ListNode head) {        if(head==null)        return null;        int len = 0;        ListNode p = head;        while(p!=null)        {            len++;            p = p.next;        }                int[] num = new int[len];        p = head;        int i = 0;        while(p!=null)        {            num[i++] = p.val;            p = p.next;        }        TreeNode root = sortedArrayToBST(num);        return root;    }    public TreeNode sortedArrayToBST(int[] num){TreeNode head = null;if(num.length==0)return null;head = creatBST(num, 0, num.length-1);return head;}public TreeNode creatBST(int[] num,int begin,int end){if(begin>end)return null;else{int mid = (begin+end)/2;TreeNode parent = new TreeNode(num[mid]);parent.left = creatBST(num, begin, mid-1);parent.right = creatBST(num, mid+1, end);return parent;}}

0 0