Convert Sorted List to Binary Search Tree

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

思路：找到中间节点，中间节点是BST树根，然后中间节点的左半部分是树根的左子树，右半部分是树根的右子树，依次递归下去。

/** * 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)     {        int nlen = 0,i=0;        ListNode p =head;        while(p!=null)        {            nlen++;            p=p.next;        }        if(nlen==0)            return null;        int[] num = new int[nlen];        while(head!=null)        {            num[i++]=head.val;            head=head.next;        }        int mid = nlen/2;        TreeNode root = new TreeNode(num[mid]);        helper(num,0,mid-1,root,0);        helper(num,mid+1,nlen-1,root,1);        return root;    }    public void helper(int[] num,int i,int j,TreeNode cur,int flag)    {        if(i<=j)        {            int l = i, r = j;            int mid = l+(r-l)/2;            TreeNode tn = new TreeNode(num[mid]);            if(flag==0) // left            {                cur.left=tn;                helper(num,l,mid-1,tn,0);                helper(num,mid+1,r,tn,1);            }            if(flag==1) // right            {                cur.right=tn;                helper(num,l,mid-1,tn,0);                helper(num,mid+1,r,tn,1);            }        }    }}

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