Convert Sorted List to Binary Search Tree

方案1：找中间结点作为root + recursively 建立左右子树

1）快慢指针法找中间结点mid，作为root,

2)  recursive problem: head 到 prev 这一段链表建立左子树， mid->next 到结尾这一段链表建立右子树

复杂度分析：T(n) = 2T(n/2) + n/2 应该是 (n/2) * lgn

TreeNode *sortedListToBST(ListNode *head) {        // write your code here        if(head==NULL) return NULL;        if(head->next==NULL) return new TreeNode(head->val);        auto p=head,q=head,preP=head;        while(q && q->next)        {            preP=p;            p=p->next;            q=q->next->next;        }        preP->next=NULL;        auto root = new TreeNode(p->val);        root->left=sortedListToBST(head);        root->right=sortedListToBST(p->next);        return root;    }

方案2：自底向上

定义一个函数 ListNode * sortedListToBST(ListNode *head, int start, int end)，含义为：将start和end对应的结点转化成二叉树，同时指针推进到end之后

1）建立左子树，同时链表指针推进过处理过的结点

2）当前结点所为root

3)  建立右子树，同时链表指针推进过处理过的结点

TreeNode *sortedListToBST(ListNode *head) {        // write your code here        int n = 0;        for (auto p = head; p; p = p->next) ++n;        return sortedListToBST(head, 0, n - 1);    }        TreeNode* sortedListToBST(ListNode* &head, int start, int end) {        if (start > end) return NULL;        int mid = start + (end - start) / 2;        auto left = sortedListToBST(head, start, mid - 1);        auto root = new TreeNode(head->val);        root->left = left;        head = head->next;        root->right = sortedListToBST(head, mid + 1, end);        return root;    }