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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.

分析：将一个升序排列的链表转换为平衡二叉搜索树，採用递归的方式，先找到链表的中点，作为二叉树的根，然后递归求解左右子树。

例如以下：

class Solution {public:    TreeNode *sortedListToBST(ListNode *head) {        if(!head){            return NULL;        }        if(!head->next){            return new TreeNode(head->val);        }        ListNode* midNode = findMidNode(head);        TreeNode* root = new TreeNode(midNode->val);        TreeNode* leftSubTree = sortedListToBST(head);        TreeNode* rightSubTree = sortedListToBST(midNode->next);        if(leftSubTree){            root->left = leftSubTree;        }        if(rightSubTree){            root->right = rightSubTree;        }        return root;    }    ListNode* findMidNode(ListNode* head){        ListNode* node1 = head;        if(!node1->next){            return node1;        }        ListNode* node2 = head->next;        while(node2 && node2->next && node2->next->next){            node2 = node2->next;            node2 = node2->next;            node1 = node1->next;        }        ListNode* midNode = node1->next;        node1->next = NULL;        return midNode;    }};

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