[LeetCode系列]有序链表转换为平衡BST的递归解法

给定有序链表(元素由小到大), 试问如何将其转换为一个平衡BST?

平衡BST: 任意节点的左右子树的深度差值不大于1.

主要思想是用递归. Trick是使用快慢指针来获取中间节点. 获得中间节点后, 将其设为此次递归的root, 随后删除此节点, 并将前一节点的next置NULL. 随后, 对中间节点的前后部分分别进行递归调用, 并将返回值作为其左右子树. 

代码如下:

 1 /**  2  * Definition for singly-linked list.  3  * struct ListNode {  4  * int val;  5  * ListNode *next;  6  * ListNode(int x) : val(x), next(NULL) {}  7  * };  8  */

 9 /** 10  * Definition for binary tree 11  * struct TreeNode { 12  * int val; 13  * TreeNode *left; 14  * TreeNode *right; 15  * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 16  * }; 17  */

18 class Solution { 19 public: 20     TreeNode *sortedListToBST(ListNode *head) { 21         if(head == NULL) return NULL; 22         if(head->next == NULL) return new TreeNode(head->val); 23         ListNode *step1 = head; 24         ListNode *step2 = head->next; 25         while(step2->next != NULL && step2->next->next != NULL){ 26             step1 = step1->next; 27             step2 = step2->next->next; 28  } 29         TreeNode *root  = new TreeNode(step1->next->val); 30         ListNode *head2 = step1->next->next; 31         delete step1->next; 32         step1->next = NULL; // cut list into two parts

33         root->left = sortedListToBST(head); 34         root->right = sortedListToBST(head2); 35         return root; 36  } 37 };