Leetcode Convert Sorted List to Binary Search Tree problem solution YASH PAL, 31 July 202419 January 2026 In this Leetcode Convert Sorted List to Binary Search Tree problem solution, we have given the head of a singly linked list where elements are sorted in ascending order, convert to a height-balanced BST. For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differs by more than 1.Leetcode Convert Sorted List to Binary Search Tree problem solution in Python.def sortedListToBST(self, head): if not head: return None return self.DFS(head) def DFS(self, head): if not head: return None if not head.next: return TreeNode(head.val) prev, slow, fast = None, head, head while fast and fast.next: prev = slow slow = slow.next fast = fast.next.next tmp = prev.next prev.next = None node = TreeNode(slow.val) node.left = self.DFS(head) node.right = self.DFS(slow.next) return node Convert Sorted List to Binary Search Tree problem solution in Java.class Solution { public TreeNode sortedListToBST(ListNode head) { if(head==null) return null; else return BST(head,null); } public static TreeNode BST(ListNode head,ListNode tail){ ListNode slow=head; ListNode fast=head; if(head==tail) return null; while(fast!=tail&&fast.next!=tail){ slow=slow.next; fast=fast.next.next; } TreeNode root=new TreeNode(slow.val); root.left=BST(head,slow); root.right=BST(slow.next,tail); return root; } } Problem solution in C++.class Solution { public: TreeNode* sortedListToBST(ListNode* head) { if(!head) return NULL; ListNode *slow=head; ListNode *fast=head->next,*tt=NULL; while(fast and fast->next) { tt=slow; slow=slow->next; fast=fast->next->next; } if(tt) tt->next=NULL; else head=NULL; TreeNode *curr=new TreeNode(slow->val); TreeNode *r=sortedListToBST(slow->next); TreeNode *l=sortedListToBST(head); curr->left=l;curr->right=r; return curr; } }; Problem solution in C.struct TreeNode* get(int v){ struct TreeNode* x; x=(struct TreeNode*)malloc(sizeof(struct TreeNode)); x->val=v; x->left=NULL; x->right=NULL; return x; } struct TreeNode* func(int *b,int l,int h){ struct TreeNode* root; int m=(l+h)/2; if(h>=l){ root=get(b[m]); root->left=func(b,l,m-1); root->right=func(b,m+1,h); return root; } else return NULL; } struct TreeNode* sortedListToBST(struct ListNode* head){ struct TreeNode* t; int a=pow(10,4); int b[2*a]; int i=0,j; while(head!=NULL){ b[i]=head->val; head=head->next; i++; } int l=0,h=i-1; t=func(b,l,h); return t; } coding problems solutions Leetcode Problems Solutions Leetcode