Unique Binary Search Trees I & II

Given n, how many structurally unique BSTs (binary search trees) that store values 1...n?

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

分析：

当数组为 1，2，3，4，.. i，.. n时，基于以下原则的BST建树具有唯一性：以i为根节点的树，其左子树由[0, i-1]构成， 其右子树由[i+1, n]构成。那么，左子树由[0 , i -1]组成时，又有多少组合方式呢？这就可以递归了嘛。

 public class Solution {
     public int numTrees(int n) {
         int[] result = new int[n + ];
         result[] = result[] = ;

         for (int i = ; i <= n; ++i) {
             for (int j = ; j < i; ++j) {  // j refers to the number of nodes in the left subtree
                 result[i] += result[j] * result[i - j - ];
             }
         }
         return result[n];
     }
 }

Unique Binary Search Trees II

Given n, generate all structurally unique BST's (binary search trees) that store values 1...n.

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

分析：

这题是要求把每一个树都找出来。从上一题我们知道，当root选取i时，它的左子树是节点从1到i -1的数组成的，右子树是i + 1到n的数组成的。那么我们创建一个函数，返回从 start 到end这两个值之间所有树的root.

 /**
  * Definition of TreeNode:
  * public class TreeNode {
  *     public int val;
  *     public TreeNode left, right;
  *     public TreeNode(int val) {
  *         this.val = val;
  *         this.left = this.right = null;
  *     }
  * }
  */
 public class Solution {
     /**
      * @paramn n: An integer
      * @return: A list of root
      */
      public ArrayList<TreeNode> generateTrees(int n) {
         return generate(, n);
     }

     private ArrayList<TreeNode> generate(int start, int end){
         ArrayList<TreeNode> treeList = new ArrayList<>();   

         if(start > end){
             treeList.add(null);
             return treeList;
         }

         for(int i = start; i <= end; i++){
             ArrayList<TreeNode> left = generate(start, i-);
             ArrayList<TreeNode> right = generate(i+, end);
             for(TreeNode l: left){
                 for(TreeNode r: right){
                     TreeNode root = new TreeNode(i);
                     root.left = l;
                     root.right = r;
                     treeList.add(root);
                 }
             }
         }
         return treeList;
     }
 }