[Geeks for Geeks] Sorted Link List to BST

Sorted Link List to BST

Given a Singly Linked List which has data members sorted in ascending order. Construct a Balanced Binary Search Tree which has same data members as the given Linked List.

Note: There might be nodes with same value.

  • Time : O(nlogn)
  • Space : O(logn)
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/* 
//Linked List
struct LNode
{
int data;
struct LNode* next;

LNode(int x){
data = x;
next = NULL;
}
};

//Tree
struct TNode
{

int data;
struct TNode* left;
struct TNode* right;
TNode(int x)
{
data=x;
left=right=NULL;
}
}; */
class Solution{
public:
TNode* sortedListToBST(LNode *head) {
if(head == nullptr) return nullptr;
if(head->next == nullptr) return new TNode(head->data);

LNode *slow = head, *fast = head;
while(fast->next->next and fast->next->next->next) {
fast = fast->next->next;
slow = slow->next;
}

LNode* mid = slow->next;
LNode* right = mid->next;

slow->next = nullptr;
mid->next = nullptr;

TNode* root = new TNode(mid->data);
root->left = sortedListToBST(head);
root->right = sortedListToBST(right);

return root;
}
};
  • Time : O(n)
  • Space : O(n)
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class Solution{
int count(LNode* head) {
LNode* runner = head;
int c = 0;
while(head) {
c++;
head = head->next;
}
return c;
}
TNode* dnc(LNode **head, int n) {
if(n <= 0) return 0;
TNode* left = dnc(head, n / 2);
TNode* root = new TNode((*head)->data);
(*head) = (*head)->next;
TNode* right = dnc(head, n - n / 2 - 1);
root->left = left;
root->right = right;
return root;
}
public:
TNode* sortedListToBST(LNode *head) {
if(!head) return nullptr;
int c = count(head);
return dnc(&head, c);
}
};
Author: Song Hayoung
Link: https://songhayoung.github.io/2022/05/19/PS/GeeksforGeeks/sorted-list-to-bst/
Copyright Notice: All articles in this blog are licensed under CC BY-NC-SA 4.0 unless stating additionally.