題目：

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

解題思路：

注意是讓構造平衡二叉搜索樹。

每次將鏈表從中間斷開，分成左右兩部分。左邊部分用來構造左子樹，右邊部分用來構造右子樹。遞歸進行求解。

代碼：

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 23 ListNode *slow = head, *fast = head, *pre = NULL; 24 do { 25 fast = fast->next; 26 if (fast == NULL) break; 27 fast = fast->next; 28 29 pre = slow; 30 slow = slow->next; 31 } while (fast != NULL); 32 33 TreeNode * root = new TreeNode(slow->val); 34 if (pre != NULL) { 35 pre->next = NULL; 36 root->left = sortedListToBST(head); 37 } 38 else { 39 root->left = NULL; 40 } 41 TreeNode * right_root = sortedListToBST(slow->next); 42 root->right = right_root; 43 if (pre != NULL) pre->next = slow; 44 return root; 45 } 46 };

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轉載于:https://www.cnblogs.com/dongguangqing/p/3728160.html

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