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 };