- Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
- Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
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 == nullptr) return nullptr;
22
23 ListNode* slow = head;
24 ListNode* fast = head;
25 ListNode* slow_pre = nullptr;
26 while(fast && fast->next){
27 slow_pre = slow;
28 slow = slow->next;
29 fast = fast->next->next;
30 }
31 if(slow_pre){
32 slow_pre->next = nullptr;
33 }
34
35 TreeNode* root = new TreeNode(slow->val);
36 if(slow!= head)
37 root->left = sortedListToBST(head);
38 else
39 root->left = nullptr;
40 root->right = sortedListToBST(slow->next);
41
42 return root;
43 }
44 };
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode *sortedArrayToBST(vector<int> &num) {
int n = num.size();
return sortedArrayToBST(num,0,n-1);
}
TreeNode* sortedArrayToBST(vector<int>& num,int l,int r){
if(l > r) return nullptr;
int mid = l + (r - l+1)/2; //加1保证左子树节点个数不少于右边的节点个数，不加1也是平衡二叉树搜索树，但是无法通过验证
TreeNode* root = new TreeNode(num[mid]);
root->left = sortedArrayToBST(num,l,mid - 1);
root->right = sortedArrayToBST(num,mid+1,r);
return root;
}
};
