diff --git a/Balanced Binary Tree Question b/Balanced Binary Tree Question
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+110. Balanced Binary Tree {Easy}
+
+Given a binary tree, determine if it is
+height-balanced.
+
+Example 1:
+Input: root = [3,9,20,null,null,15,7]
+Output: true
+
+Example 2:
+Input: root = [1,2,2,3,3,null,null,4,4]
+Output: false
+
+Example 3:
+Input: root = []
+Output: true
+
+Constraints:
+
+The number of nodes in the tree is in the range [0, 5000].
+-104 <= Node.val <= 104
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diff --git a/BalancedBinaryTree.java b/BalancedBinaryTree.java
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+/*
+Time Complexity:
+The method height is a recursive function that visits each node once.
+At each node:
+    It recursively computes the height of the left and right subtrees.
+    It performs constant-time checks (e.g., if conditions, Math.abs, and Math.max).
+Thus, the total time complexity is O(n), where n is the number of nodes in the tree.
+
+Space Complexity:
+The space complexity is O(h), where h is the height of the tree:
+    Worst Case (Skewed Tree): O(n)
+    Best Case (Balanced Tree): O(logn)
+
+Did this code successfully run on Leetcode: Yes
+*/
+
+//Definition for a binary tree node.
+class TreeNode {
+      int val;
+      TreeNode left;
+      TreeNode right;
+      TreeNode() {}
+      TreeNode(int val) { this.val = val; }
+      TreeNode(int val, TreeNode left, TreeNode right) {
+          this.val = val;
+          this.left = left;
+          this.right = right;
+      }
+ }
+
+class BalancedBinaryTree {
+    public boolean isBalanced(TreeNode root)
+    {
+        if(root == null) return true;
+        return height(root) != -1;
+    }
+
+    private int height(TreeNode root)
+    {
+        //base
+        if(root == null) return 0;
+
+        //logic
+        int left = height(root.left);
+        int right = height(root.right);
+
+        if(left == -1 || right == -1) return -1;
+        if(Math.abs(left-right) > 1) return -1;
+
+        return Math.max(left, right) + 1;
+    }
+}
diff --git a/Palindrome Linked List Question b/Palindrome Linked List Question
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+++ b/Palindrome Linked List Question	
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+234. Palindrome Linked List {Easy}
+
+Given the head of a singly linked list, return true if it is a palindrome or false otherwise.
+
+Example 1:
+
+Input: head = [1,2,2,1]
+Output: true
+
+Example 2:
+
+Input: head = [1,2]
+Output: false
+
+
+Constraints:
+
+The number of nodes in the list is in the range [1, 105].
+0 <= Node.val <= 9
+
+
+Follow up: Could you do it in O(n) time and O(1) space?
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diff --git a/PalindromeLinkedList.java b/PalindromeLinkedList.java
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+++ b/PalindromeLinkedList.java
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+/*
+Time complexity : O(n), where n is the number of nodes in the Linked List.
+Finding the middle is O(n), reversing a list in place is O(n), and then comparing the 2 resulting Linked Lists
+is also O(n) => O(n) + O(n) + O(n) = O(3n) = O(n)
+
+Space complexity : O(1) as we are not using any auxiliary data structure
+
+Did this code successfully run on Leetcode: Yes
+*/
+
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode() {}
+ *     ListNode(int val) { this.val = val; }
+ *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
+ * }
+ */
+class PalindromeLinkedList {
+    public boolean isPalindrome(ListNode head)
+    {
+        //base case
+        if(head == null) return true;
+
+        //logic
+        ListNode slow = head;
+        ListNode fast = head;
+
+        //Finding the mid of the list
+        while(fast.next != null && fast.next.next != null)
+        {
+            slow = slow.next;
+            fast = fast.next.next;
+        }
+
+        //Now we will be having the mid point at slow
+        ListNode head2 = slow.next; //slow.next here is the midpoint
+        slow.next = null;
+
+        /*
+          1 -> 2 -> 2 -> 3 -> 2 -> 2 -> 1 -> null (3 is midpoint which is also slow)
+
+          After finding the midpoint
+          left sublist  = 1 -> 2 -> 2 -> 3 -> null (3 is head2)
+          right sublist = 2 -> 2 -> 1 -> null
+        */
+
+        /*
+            Now we need to reverse the right sublist to compare it with left sublist, make head2 point to the reversed list first node
+            meaning now it is pointing 2 make it point to 1
+        */
+        head2 = reverseList(head2);
+
+        //After getting the reverse list, check whether elements in both the list are equal
+        while(head != null && head2 != null)
+        {
+            if(head.val != head2.val) {
+                return false;
+            }
+
+            head = head.next;
+            head2 = head2.next;
+        }
+
+        return true;
+    }
+
+    private ListNode reverseList(ListNode head)
+    {
+        ListNode prev = null;
+        ListNode curr = head;
+
+        while(curr != null)
+        {
+            ListNode nextTemp = curr.next;
+            curr.next = prev;
+            prev = curr;
+            curr = nextTemp;
+        }
+
+        return prev;
+    }
+}
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diff --git a/Reverse Linked List Question b/Reverse Linked List Question
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+https://leetcode.com/problems/reverse-linked-list/description/
+
+206. Reverse Linked List {Easy}
+
+Given the head of a singly linked list, reverse the list, and return the reversed list.
+
+Example 1:
+
+Input: head = [1,2,3,4,5]
+Output: [5,4,3,2,1]
+Example 2:
+
+Input: head = [1,2]
+Output: [2,1]
+Example 3:
+
+Input: head = []
+Output: []
+
+Constraints:
+
+The number of nodes in the list is the range [0, 5000].
+-5000 <= Node.val <= 5000
+
+
+Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both?
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diff --git a/ReverseLinkedListIterative.java b/ReverseLinkedListIterative.java
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+++ b/ReverseLinkedListIterative.java
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+/*
+Time complexity : O(n). List of n nodes, traverse over each node once
+
+Space complexity : O(1).
+
+Did this code successfully run on Leetcode: Yes
+*/
+
+//Definition for singly-linked list.
+class ListNode {
+      int val;
+      ListNode next;
+      ListNode() {}
+      ListNode(int val) { this.val = val; }
+      ListNode(int val, ListNode next) { this.val = val; this.next = next; }
+ }
+
+class ReverseLinkedListIterative {
+    public ListNode reverseList(ListNode head)
+    {
+        ListNode prev = null;
+        ListNode curr = head;
+
+        while(curr != null)
+        {
+            ListNode nextTemp = curr.next;
+            curr.next = prev;
+            prev = curr;
+            curr = nextTemp;
+        }
+
+        return prev;
+    }
+}
diff --git a/ReverseLinkedListRecursive.java b/ReverseLinkedListRecursive.java
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+++ b/ReverseLinkedListRecursive.java
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+/*
+Time complexity : O(n). List of n nodes, traverse over each node once
+
+Space complexity : O(n)
+The extra space comes from implicit stack space due to recursion. The recursion could go up to n levels deep. Means,
+so list of length n we make n recursive calls.
+
+Did this code successfully run on Leetcode: Yes
+*/
+
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode() {}
+ *     ListNode(int val) { this.val = val; }
+ *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
+ * }
+ */
+class ReverseLinkedListRecursive {
+    public ListNode reverseList(ListNode head)
+    {
+        //base case
+        if( head == null || head.next == null) return head;
+
+        //logic
+        ListNode reverseSubListHead = reverseList(head.next);
+        head.next.next = head;
+        head.next = null;
+
+        return reverseSubListHead;
+    }
+}
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