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：把一个排好序的链表转化为平衡二叉树。这道题主要是递归做法，每次找到中间结点，将其作为根结点，然后递归左右子链表，依次找到左右直接点，这样就可以建造一棵平衡二叉树了。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int ListLength(ListNode *head)
    {
        int len=0;
        while(head)
        {
            head=head->next;
            len++;
        }
        return len;
    }
    TreeNode *createBST(ListNode *head,int left,int right)
    {
        if(left>right)
            return NULL;
        int mid=(left+right)>>1;
        ListNode *cur=head;
        for(int i=left;i<mid;i++)
            cur=cur->next;
        TreeNode *leftTree=createBST(head,left,mid-1);
        TreeNode *rightTree=createBST(cur->next,mid+1,right);
        TreeNode *root=new TreeNode(cur->val);
        root->left=leftTree;
        root->right=rightTree;
        return root;
    }
    TreeNode *sortedListToBST(ListNode *head) {
        if(head==NULL)
            return NULL;
        int len;
        len=ListLength(head);
        return createBST(head,0,len-1);
    }
};