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+
+/*
+https://leetcode.com/problems/unique-binary-search-trees/description/
+
+96. Unique Binary Search Trees
+
+Given n, how many structurally unique BST's (binary search trees) that store values 1 ... n?
+
+Example:
+
+Input: 3
+Output: 5
+Explanation:
+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
+   
+
+DP Solution: https://www.programcreek.com/2014/05/leetcode-unique-binary-search-trees-java/
+*/
+
+// Solution 3 using DP
+var numTrees3 = function (n) {
+  if (n == 0)
+    return 0
+
+  var map = [];
+  map[0] = 1;
+  map[1] = 1;
+
+  for(var i = 2; i <= n; i++) {
+    var currentI = 0;
+    for(var j = 0; j < i; j++) {
+      currentI += map[j] * map[i - j - 1];
+    }
+    map[i] = currentI;
+  }
+
+  return map[n];
+}
+
+// Solution 2 (Solution 1 + Memoization)
+var numTrees2 = function(n) {
+  var memo = {};
+  return numTreesAux2(1, n, memo);
+};
+
+var numTreesAux2 = function(leftMin, leftMax, memo) {
+  const keyMemo = buildKey(leftMin, leftMax);
+  if(memo[keyMemo]) {
+    return memo[keyMemo] 
+  }
+
+  if(leftMin > leftMax)
+      return 0;
+  
+  if(leftMin === leftMax) 
+      return 1;
+  
+  var count = 0;
+  for(var i = leftMin; i <= leftMax; i++){  
+      const left = numTreesAux2(leftMin, i - 1, memo);
+      const right = numTreesAux2(i + 1, leftMax, memo);  
+      
+      if(left > 0 && right > 0) {
+          count += left * right; 
+      } else {
+          count += (left > 0) ? left : right;
+      }
+  } 
+  
+  memo[keyMemo] = count;
+  return count;
+}
+
+var buildKey = function(a, b) {
+  return a + "-" + b;
+}
+
+
+// Solution 1
+var numTrees1 = function(n) {
+  return numTreesAux1(1, n);
+};
+
+var numTreesAux1 = function(leftMin, leftMax) {
+  if(leftMin > leftMax)
+      return 0;
+  
+  if(leftMin === leftMax) 
+      return 1;
+  
+  var count = 0;
+  for(var i = leftMin; i <= leftMax; i++){  
+      const left = numTreesAux1(leftMin, i - 1);
+      const right = numTreesAux1(i + 1, leftMax);  
+      
+      if(left > 0 && right > 0) {
+          count += left * right; 
+      } else {
+          count += (left > 0) ? left : right;
+      }
+  } 
+  
+  return count;
+}
+
+var main = function() {
+  console.log(numTrees1(1));
+  console.log(numTrees1(2));
+  console.log(numTrees1(3));
+  console.log(numTrees1(5));
+
+  console.log(numTrees2(1));
+  console.log(numTrees2(2));
+  console.log(numTrees2(3));
+  console.log(numTrees2(5));
+
+  console.log(numTrees3(1));
+  console.log(numTrees3(2));
+  console.log(numTrees3(3));
+  console.log(numTrees3(5));
+}
+
+main();
+
+module.exports.main = main