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，采用递归的方式，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。

如下：

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;
    }
};