LINTCODE---排序列表转换为二分查找树

LINTCODE—排序列表转换为二分查找树

思路：主要就是用快慢指针找出中间点，作为一个伪“顶点”，然后递归调用自身，返回左右节点；

class Solution {public:    /*     * @param head: The first node of linked list.     * @return: a tree node     */    TreeNode * sortedListToBST(ListNode * head) {        // write your code here        if(head == NULL )            return NULL;        if(head -> next == NULL)            return new TreeNode(head -> val);        //快慢指针找到中间点        ListNode *slow = head ,*fast = head ,*temp = head;        while(fast != NULL && fast->next != NULL)        {            temp = slow;            slow = slow -> next;            fast = fast -> next -> next;        }        temp -> next = NULL;        // 递归调用自身        TreeNode *resTree = new TreeNode(slow -> val);        resTree -> left = sortedListToBST(head);        resTree -> right = sortedListToBST(slow -> next);        return resTree;    }};