diff --git a/Nth-node_of_inorder_traversal.cpp b/Nth-node_of_inorder_traversal.cpp
new file mode 100644
index 0000000..5c4307c
--- /dev/null
+++ b/Nth-node_of_inorder_traversal.cpp
@@ -0,0 +1,63 @@
+// C program for nth nodes of inorder traversals
+#include <stdio.h>
+#include <stdlib.h>
+
+/* A binary tree node has data, pointer to left child
+and a pointer to right child */
+struct Node {
+	int data;
+	struct Node* left;
+	struct Node* right;
+};
+
+/* Helper function that allocates a new node with the
+given data and NULL left and right pointers. */
+struct Node* newNode(int data)
+{
+	struct Node* node =
+		(struct Node*)malloc(sizeof(struct Node));
+	node->data = data;
+	node->left = NULL;
+	node->right = NULL;
+
+	return (node);
+}
+
+/* Given a binary tree, print its nth nodes of inorder*/
+void NthInorder(struct Node* node, int n)
+{
+	static int count = 0;
+	if (node == NULL)
+		return;
+
+	if (count <= n) {
+
+		/* first recur on left child */
+		NthInorder(node->left, n);
+		count++;
+
+		// when count = n then print element
+		if (count == n)
+			printf("%d ", node->data);
+
+		/* now recur on right child */
+		NthInorder(node->right, n);
+	}
+}
+
+/* Driver program to test above functions*/
+int main()
+{
+	struct Node* root = newNode(10);
+	root->left = newNode(20);
+	root->right = newNode(30);
+	root->left->left = newNode(40);
+	root->left->right = newNode(50);
+
+	int n = 4;
+
+	NthInorder(root, n);
+	return 0;
+}
+
+//Time Complexity: O(n)