Convert Sorted List to Binary Search Tree
Problem Statement :
Given the head of a singly linked list where elements are sorted in ascending order, convert it to a height-balanced binary search tree. Example 1: Input: head = [-10,-3,0,5,9] Output: [0,-3,9,-10,null,5] Explanation: One possible answer is [0,-3,9,-10,null,5], which represents the shown height balanced BST. Example 2: Input: head = [] Output: [] Constraints: The number of nodes in head is in the range [0, 2 * 104]. -105 <= Node.val <= 105
Solution :
Solution in C :
int listLenth(struct ListNode* head){
struct ListNode* temp = head;
int cn = 0;
while(temp){
cn++;
temp = temp->next;
}
return cn;
}
struct TreeNode* sortedArrayToBST(int* nums, int n){
if(n <= 0)
return NULL;
struct TreeNode *root = malloc( sizeof(struct TreeNode));
if(n == 1){
root->val = nums[0];
root->left = NULL;
root->right = NULL;
}
else{
root->val = nums[n/2];
root->left = sortedArrayToBST(nums, n/2);
root->right = sortedArrayToBST(&nums[n/2 + 1], n - 1 - n/2);
}
return root;
}
struct TreeNode* sortedListToBST(struct ListNode* head){
int n = listLenth(head);
if (n == 0)
return NULL;
int* nums = malloc(n * sizeof(int));
struct ListNode* temp = head;
for(int i = 0; i < n; i++){
nums[i] = temp->val;
temp = temp->next;
}
return sortedArrayToBST(nums, n);
}
Solution in C++ :
class Solution {
public:
ListNode* middleNode(ListNode* head) {
if(!head || !head->next)
return head;
ListNode* s=head;
ListNode* f=head;
ListNode * p= NULL;
while(f && f->next)
{
p=s;
s=s->next;
f=f->next->next;
}
return p;
}
TreeNode* sortedListToBST(ListNode* head) {
if(!head)
return NULL;
if(!head->next)
return new TreeNode(head->val);
ListNode* n=middleNode(head);
ListNode* tmp=n->next;
n->next=NULL;
TreeNode * t=new TreeNode(tmp->val);
t->left = sortedListToBST(head);
t->right = sortedListToBST(tmp->next);
return t;
}
};
Solution in Java :
class Solution {
public TreeNode sortedListToBST(ListNode head) {
if(head == null)
{
return null;
}
return createBST(head,null);
}
public TreeNode createBST(ListNode start,ListNode end)
{
if(start==end)return null;
ListNode mid = middle(start,end);
TreeNode root = new TreeNode(mid.val);
root.left = createBST(start,mid);
root.right = createBST(mid.next,end);
return root;
}
public ListNode middle(ListNode head,ListNode end)
{
ListNode slow = head;
ListNode fast = head;
while(fast!= end && fast.next != end)
{
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
}
Solution in Python :
class Solution:
def sortedListToBST(self, head: ListNode) -> TreeNode:
if not head:
return None
if not head.next:
return TreeNode(head.val)
slow, fast = head, head.next.next
while fast and fast.next:
slow = slow.next
fast = fast.next.next
root = TreeNode(slow.next.val)
root.right = self.sortedListToBST(slow.next.next)
slow.next = None
root.left = self.sortedListToBST(head)
return root
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