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);

    }

};