diff --git a/cpp/096_Unique_Binary_Search_Trees.cpp b/cpp/096_Unique_Binary_Search_Trees.cpp
new file mode 100644
index 0000000..5705462
--- /dev/null
+++ b/cpp/096_Unique_Binary_Search_Trees.cpp
@@ -0,0 +1,54 @@
+// 96. Unique Binary Search Trees
+/**
+ * Given n, how many structurally unique BST's (binary search trees) that store values 1...n?
+ * 
+ * For example,
+ * Given n = 3, there are a total of 5 unique BST's.
+ * 
+ *    1         3     3      2      1
+ *     \       /     /      / \      \
+ *      3     2     1      1   3      2
+ *     /     /       \                 \
+ *    2     1         2                 3
+ * 
+ * Tags: Tree, Dynamic Programming
+ * 
+ * Similar Problems: (M) Unique Binary Search Trees II
+ * 
+ * Author: Kuang Qin
+ */
+
+#include <iostream>
+
+using namespace std;
+
+// G(n):    the number of unique BST for a sequence of length n (G(0) = G(1) = 1)
+// F(i, n): the number of unique BST, where the number i is the root of BST (1 <= i <= n)
+// G(n) = F(1, n) + F(2, n) + ... + F(n, n)
+// F(i, n) = G(i - 1) * G(n - i): 
+// for example, F(3, 7): 3 as root, [1, 2] left subtree, [4, 5, 6, 7] right subtree
+// F(3, 7) = G(2) * G(4)
+// G(n) = G(0) * G(n - 1) + G(1) * G(n - 2) + ... + G(n - 1) * G(0)
+class Solution {
+public:
+    int numTrees(int n) {
+        int G[n + 1] = {};
+        G[0] = G[1] = 1;
+        
+        for (int i = 2; i <= n; i++) {
+            for (int j = 1; j <= i; j++) {
+                G[i] += G[j - 1] * G[i - j];
+            }
+        }
+        
+        return G[n];
+    }
+};
+
+int main() {
+    Solution sol;
+    int num = sol.numTrees(3);
+    cout << num << endl;
+    cin.get();
+    return 0;
+}
\ No newline at end of file