96

Author: lzl124631x

leetcode/96. Unique Binary Search Trees/README.md

@@ -0,0 +1,66 @@
+# [96. Unique Binary Search Trees (Medium)](https://leetcode.com/problems/unique-binary-search-trees/)
+
+<p>Given <em>n</em>, how many structurally unique <strong>BST's</strong> (binary search trees) that store values 1 ...&nbsp;<em>n</em>?</p>
+
+<p><strong>Example:</strong></p>
+
+<pre><strong>Input:</strong> 3
+<strong>Output:</strong> 5
+<strong>Explanation:
+</strong>Given <em>n</em> = 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
+</pre>
+
+
+**Related Topics**:  
+[Dynamic Programming](https://leetcode.com/tag/dynamic-programming/), [Tree](https://leetcode.com/tag/tree/)
+
+**Similar Questions**:
+* [Unique Binary Search Trees II (Medium)](https://leetcode.com/problems/unique-binary-search-trees-ii/)
+
+## Solution 1. DP
+
+Let `dp[i]` be the answer.
+
+To get `dp[i]`, we can pick `j` as the root node (`1 <= j <= i`), then there are `j - 1` nodes in the left sub-tree and `i - j` nodes in the right sub-tree, and they have `dp[j - 1]` and `dp[i - j]` unique BSTs respectively. So `dp[i] = sum( dp[j - 1] * dp[i - j] | 1 <= j <= i)`
+
+```cpp
+// OJ: https://leetcode.com/problems/unique-binary-search-trees
+// Author: github.com/lzl124631x
+// Time: O(N^2)
+// Space: O(N)
+class Solution {
+public:
+    int numTrees(int n) {
+        vector<int> dp(n + 1);
+        dp[0] = 1;
+        for (int i = 1; i <= n; ++i) {
+            for (int j = 1; j <= i; ++j) dp[i] += dp[j - 1] * dp[i - j];
+        }
+        return dp[n];
+    }
+};
+```
+
+## Solution 2. Catalan Number
+
+```cpp
+// OJ: https://leetcode.com/problems/unique-binary-search-trees
+// Author: github.com/lzl124631x
+// Time: O(N)
+// Space: O(1)
+// Ref: https://discuss.leetcode.com/topic/13321/a-very-simple-and-straight-ans-based-on-math-catalan-number-o-n-times-o-1-space
+class Solution {
+public:
+    int numTrees(int n) {
+        long long ans = 1, i;
+        for (i = 1; i <= n; ++i) ans = ans * (i + n) / i;
+        return ans / i;
+    }
+};
+```