diff --git a/Problem1.cs b/Problem1.cs
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+++ b/Problem1.cs
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+// Time Complexity : O(n)
+// Space Complexity : O(n) to store the dictionary which contains the mapping for inorder array
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No
+
+
+// Your code here along with comments explaining your approach
+
+/*
+I use the postorder list to keep track of values of root nodes (Postorder = Recursively traverse left subtree -> Recursively traverse right subtree -> Visit root) and inorder list to obtain information about
+    what nodes belong to the left subtree of a given node and what nodes belong to the right subtree of a given node. 
+*/
+
+
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     public int val;
+ *     public TreeNode left;
+ *     public TreeNode right;
+ *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
+ *         this.val = val;
+ *         this.left = left;
+ *         this.right = right;
+ *     }
+ * }
+ */
+public class Solution
+{
+    private Dictionary<int, int> inorderLookup;
+    private int index;
+    public TreeNode BuildTree(int[] inorder, int[] postorder)
+    {
+        inorderLookup = new();
+        index = postorder.Length - 1;
+
+        for (int i = 0; i < inorder.Length; i++)
+        {
+            inorderLookup[inorder[i]] = i;
+        }
+
+        return Helper(0, inorder.Length - 1, postorder);
+    }
+
+    public TreeNode Helper(int st, int end, int[] postorder)
+    {
+        if (st > end)
+        {
+            return null;
+        }
+
+        int rootVal = postorder[index];
+        index -= 1;
+        int rootIndex = inorderLookup[rootVal];
+        TreeNode temp = new TreeNode(rootVal);
+        temp.right = Helper(rootIndex + 1, end, postorder);
+        temp.left = Helper(st, rootIndex - 1, postorder);
+
+        return temp;
+
+    }
+}
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diff --git a/Problem2.cs b/Problem2.cs
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+++ b/Problem2.cs
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+// Time Complexity : O(n)
+// Space Complexity : O(h) space taken up by the recursion stack
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No
+
+
+// Your code here along with comments explaining your approach
+
+/*
+I perform DFS on the tree, I use a global variable totalSum to keep track of sum so far. If root is null, I simply return from the helper method. I calculate sum as sum = sum * 10 + root.val. 
+If the current node is a lead node, I add sum to totalSum and return from the helper method.
+*/
+
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     public int val;
+ *     public TreeNode left;
+ *     public TreeNode right;
+ *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
+ *         this.val = val;
+ *         this.left = left;
+ *         this.right = right;
+ *     }
+ * }
+ */
+public class Solution
+{
+    private int totalSum;
+    public int SumNumbers(TreeNode root)
+    {
+        totalSum = 0;
+        Helper(root, 0);
+        return totalSum;
+    }
+
+    private void Helper(TreeNode root, int sum)
+    {
+        if (root == null)
+        {
+            return;
+        }
+
+        sum = sum * 10 + root.val;
+
+        if (root.left == null && root.right == null)
+        {
+            totalSum += sum;
+            return;
+        }
+
+        Helper(root.left, sum);
+        Helper(root.right, sum);
+    }
+}
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