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 + 1];
        result[0] = result[1] = 1;

        for (int i = 2; i <= n; ++i) {
            for (int j = 0; j < i; ++j) {  // j refers to the number of nodes in the left subtree
                result[i] += result[j] * result[i - j - 1];
            }
        }
        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(1, 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-1);
            ArrayList<TreeNode> right = generate(i+1, 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;
    }
}