Unique Binary Search Trees I &&II

Change-Id: Id843060bbea224d18311f6e0f712cf110c8f826a

Author: applewjg

UniqueBinarySearchTrees.java

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+/*
+ Author:     King, [email protected]
+ Date:       Oct 07, 2014
+ Problem:    Unique Binary Search Trees
+ Difficulty: Easy
+ Source:     https://oj.leetcode.com/problems/unique-binary-search-trees/
+ Notes:
+ 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
+
+ Solution: dp.
+*/
+public class Solution {
+    public int numTrees_1(int n) {
+        int[] dp = new int[n+1];
+        dp[0] = 1;
+        for (int i = 1; i <= n; ++i)
+            for (int j = 0; j < i; j++)
+                dp[i] += dp[j] * dp[i-j-1];
+        return dp[n];
+    }
+    public int numTrees(int n) {
+        if (n < 0) return 0;
+        int[] dp = new int[n+1];
+        dp[0] = 1; dp[1] = 1;
+        for(int i = 2;i <= n; ++i){
+            dp[i] = dp[i-1] * (4 * i - 2)/(i + 1);
+        }
+        return dp[n];
+    }
+}

UniqueBinarySearchTreesII.java

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+/*
+ Author:     King, [email protected]
+ Date:       Dec 25, 2014
+ Problem:    Unique Binary Search Trees II
+ Difficulty: Medium
+ Source:     https://oj.leetcode.com/problems/unique-binary-search-trees-ii/
+ Notes:
+ Given 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
+
+ Solution: 1. DFS directly. (from the Internet)
+           2. DP + DFS. (my solution)
+              a. Generate trees for 'n' from 1 to n. (DP)
+              b. When generate trees for n = i, get the left and right subtrees 
+                 by copying tree structures of dp[1...i-1]. (copy tree uses DFS)
+*/
+/**
+ * Definition for binary tree
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; left = null; right = null; }
+ * }
+ */
+public class Solution {
+    public List<TreeNode> generateTrees(int n) {
+        return generateTreesRe(1, n);
+    }
+    public List<TreeNode> generateTreesRe(int l, int r) {
+        ArrayList<TreeNode> res = new ArrayList<TreeNode>();
+        if (l > r) {
+            res.add(null);
+            return res;
+        }
+        for (int k = l; k <= r; ++k) {
+            List<TreeNode> leftTrees = generateTreesRe(l, k-1);
+            List<TreeNode> rightTrees = generateTreesRe(k+1, r);
+            for (int i = 0; i < leftTrees.size(); i++) {
+                for (int j = 0; j < rightTrees.size(); j++) {
+                    TreeNode root = new TreeNode(k);
+                    root.left = leftTrees.get(i);
+                    root.right = rightTrees.get(j);
+                    res.add(root);
+                }
+            }
+        }
+        return res;
+    }
+}