[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 };

 

posted @ 2014-08-28 10:23  Lancelod_Liu  阅读(200)  评论(0)  编辑  收藏  举报