[LeetCode] 96. Unique Binary Search Trees I & 95. II

Unique Binary Search Trees

Problem

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

Note

DP

Solution

public class Solution {
    public int numTrees(int n) {
        int[] count = new int[n+1];
        count[0] = 1;
        for (int i = 1; i <= n; i++) {
            for (int root = 1; root <= i; root++) {
                int left = count[root-1];
                int right = count[i-root];
                count[i] += left*right;
            }
        }
        return count[n];
    }
}

Unique Binary Search Trees II

Problem

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

Note

DFS

Solution

class Solution {
    public List<TreeNode> generateTrees(int n) {
        if (n == 0) return new ArrayList<>();
        return helper(1, n);
    }
    private List<TreeNode> helper(int start, int end) {
        List<TreeNode> res = new ArrayList<>();
        if (start > end) {
            res.add(null);
            return res;
        }
        for (int i = start; i <= end; i++) {
            List<TreeNode> left = helper(start, i-1);
            List<TreeNode> right = helper(i+1, end);
            for (TreeNode leftNode: left) {
                for (TreeNode rightNode: right) {
                    TreeNode root = new TreeNode(i);
                    root.left = leftNode;
                    root.right = rightNode;
                    res.add(root);
                }
            }
        }
        return res;
    }
}