【leetcode】Convert Sorted List to Binary Search Tree （middle）

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

 

思路：题目看上去好像很难，但实际上很简单，递归做就行，每次找到左右子树对应的子链表就行。一次AC。

class Solution {

public:

    TreeNode *sortedListToBST(ListNode *head) {

        if(head == NULL)

            return NULL;



        ListNode * fast = head, *slow = head, *slowpre = head; //分别是快指针、慢指针、慢指针前一个指针 慢指针的位置就是当前平衡树根节点的位置 中间值

        while(fast != NULL && fast->next != NULL)

        {

            fast = fast->next->next;

            slowpre = slow;

            slow = slow->next;    

        }

        TreeNode * root = new TreeNode(slow->val);

        ListNode * left = (slow == head) ? NULL : head; //如果慢指针==头指针 则其左子树是空的

        ListNode * right = slow->next;

        slowpre->next = NULL; //左子树对应的链表末尾置null

        root->left = sortedListToBST(left);

        root->right = sortedListToBST(right);



        return root;

    }

};