diff --git a/cpp/095_Unique_Binary_Search_Trees_II.cpp b/cpp/095_Unique_Binary_Search_Trees_II.cpp
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+// 95. Unique Binary Search Trees II
+/**
+ * Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1...n.
+ * 
+ * For example,
+ * Given n = 3, your program should return all 5 unique BST's shown below.
+ * 
+ *    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, (M) Different Ways to Add Parentheses
+ * 
+ * Author: Kuang Qin
+ */ 
+
+#include <iostream>
+#include <vector>
+#include <queue>
+#include <sstream>
+
+using namespace std;
+
+/**
+ * Definition for a binary tree node.
+ */
+struct TreeNode {
+    int val;
+    TreeNode *left;
+    TreeNode *right;
+    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+};
+
+// 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 {
+    vector<TreeNode*> generate(int start, int end) {
+        vector<TreeNode*> res;
+        if (start > end) {
+            res.push_back(NULL);    // empty tree
+            return res;
+        }
+        
+        for (int iroot = start; iroot <= end; iroot++) {    // root position
+            vector<TreeNode*> left = generate(start, iroot - 1);
+            vector<TreeNode*> right = generate(iroot + 1, end);
+            
+            for (int i = 0; i < left.size(); i++) {         // left subtree
+                for (int j = 0; j < right.size(); j++) {    // right subtree
+                    TreeNode *root = new TreeNode(iroot);
+                    root->left = left[i];
+                    root->right = right[j];
+                    res.push_back(root);
+                }
+            }
+        }
+        
+        return res;
+    }
+public:
+    vector<TreeNode*> generateTrees(int n) {
+        vector<TreeNode*> res;
+        if (n == 0) {
+            return res;
+        }
+        
+        return generate(1, n);
+    }
+};
+
+class TreeOperation {
+    int getTreeHeight(TreeNode* root) {
+        if (root == NULL) {
+            return 0;
+        }
+        
+        int l = getTreeHeight(root->left);
+        int r = getTreeHeight(root->right);
+        
+        if (l > r) {
+            return l + 1;
+        }
+        
+        return r + 1;
+    }
+    
+    // output all the node in level order, including null pointers
+    vector<vector<string>> levelOrderFull(TreeNode* root) {
+        vector<vector<string>> output;
+        if (root == NULL) {
+            return output;
+        }
+        
+        int h = getTreeHeight(root);
+        queue<TreeNode*> q;
+        q.push(root);
+        
+        // fill the container in each level
+        for (int i = 0; i < h; i++) {
+            vector<string> level;
+            int currLevelCount = q.size();
+            
+            // while loop for current level
+            while (currLevelCount--) {
+                TreeNode *curr = q.front();
+                if (curr == NULL) {
+                    level.push_back("&");
+                    q.push(NULL);
+                    q.push(NULL);
+                }
+                else {
+                    stringstream ss;
+                    ss << curr->val;
+                    level.push_back(ss.str());
+                    q.push(curr->left);
+                    q.push(curr->right);
+                }
+                
+                q.pop();
+            }
+            
+            output.push_back(level);
+        }
+        
+        return output;
+    }
+public:
+    void printTree(TreeNode *root) {
+        vector<vector<string>> output = levelOrderFull(root);
+        int h = output.size();
+        if (h == 0) {
+            return;
+        }
+        
+        int w = 2 * output[h - 1].size() + 1;   // total width
+        for (int i = 0; i < h; i++) {
+            int n = output[i].size();
+            int m = (w - n) / (n + 1);  // calculate space width
+            string sp(m, ' ');
+            if ((w - n) % (n + 1)) {
+                cout << sp << " ";
+            }
+            else {
+                cout << sp;     // add space
+            }
+            
+            for (int j = 0; j < n; j++) {
+                cout << output[i][j] << sp;
+            }
+            cout << endl;
+        }
+        
+        return;
+    }
+    
+    void deleteTree(TreeNode *root) {
+        if (root == NULL) {
+            return;
+        }
+        
+        if (root->left == NULL && root->right == NULL) {
+            delete root;
+            return;
+        }
+        
+        deleteTree(root->left);
+        deleteTree(root->right);
+        return;
+    }
+};
+
+int main() {
+    Solution sol;
+    vector<TreeNode*> trees = sol.generateTrees(3);
+    TreeOperation trOp;
+    for (int i = 0; i < trees.size(); i++) {
+        trOp.printTree(trees[i]);
+        trOp.deleteTree(trees[i]);
+        cout << endl;
+    }
+    cin.get();
+    return 0;
+}
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