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In this post, you will find the solution for the **Convert Sorted List to Binary Search Tree** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Convert Sorted List to Binary Search Tree– LeetCode Problem

Convert Sorted List to Binary Search Tree– LeetCode Problem

**Problem:**

Given the `head`

of a singly linked list where elements are **sorted in ascending order**, convert it 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 differ by more than 1.

**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 * 10`

.^{4}] `-10`

^{5}<= Node.val <= 10^{5}

Convert Sorted List to Binary Search Tree– LeetCode Solutions

Convert Sorted List to Binary Search Tree Solution in C++:

class Solution { public: TreeNode* sortedListToBST(ListNode* head) { vector<int> A; // construct the array for (ListNode* curr = head; curr; curr = curr->next) A.push_back(curr->val); return helper(A, 0, A.size() - 1); } private: TreeNode* helper(const vector<int>& A, int l, int r) { if (l > r) return nullptr; const int m = l + (r - l) / 2; TreeNode* root = new TreeNode(A[m]); root->left = helper(A, l, m - 1); root->right = helper(A, m + 1, r); return root; } };

Convert Sorted List to Binary Search Tree Solution in Java:

class Solution { public TreeNode sortedListToBST(ListNode head) { List<Integer> A = new ArrayList<>(); // construct the array for (ListNode curr = head; curr != null; curr = curr.next) A.add(curr.val); return helper(A, 0, A.size() - 1); } private TreeNode helper(List<Integer> A, int l, int r) { if (l > r) return null; final int m = l + (r - l) / 2; TreeNode root = new TreeNode(A.get(m)); root.left = helper(A, l, m - 1); root.right = helper(A, m + 1, r); return root; } }

Convert Sorted List to Binary Search Tree Solution in Python:

class Solution: def sortedListToBST(self, head: ListNode) -> TreeNode: def findMid(head: ListNode) -> ListNode: prev = None slow = head fast = head while fast and fast.next: prev = slow slow = slow.next fast = fast.next.next prev.next = None return slow if not head: return None if not head.next: return TreeNode(head.val) mid = findMid(head) root = TreeNode(mid.val) root.left = self.sortedListToBST(head) root.right = self.sortedListToBST(mid.next) return root