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

对于这种找解的个数的题目，可以用DP来做。
这道题用DP的思路是，建立count[]数组，存放从1到n每一个数对应的BST的个数。
对于数n，可能有多少个BST呢？这个是由它的根节点分布决定的，若根节点为root，左子树的个数就是count[root-1]的值，右子树的个数就是count[i-root]的值，两个解的乘积就是根节点为root，借点总数为i的所有BST的个数。将root从1到i遍历一遍，就得到了count[i]，即i个结点的所有解。DP到count[n]即为所求n个结点的所有BST总数。

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

首先，重新构建class。我们在helper()函数中增加start和end两个参数，以便更好地使用递归操作。和Unique BST I一样，仍然考虑root从1到n，每种情况下限定左右子树，以对res进行更新。
注意两点：

1. 限定左右子树的写法：对于root.val = i，左右子树分别来自于集合helper(start, i-1)和helper(i+1, end)，要用Enhanced for loop进行遍历。

2. 在两个for循环内部，左右子树都已被确定，此时要将它们与root进行连接，其实并不需要copy()函数来复制已经选择好的左右子树nodeLeft和nodeRight。

Solution

public class Solution {
    public List<TreeNode> generateTrees(int n) {
        return helper(1, n);
    }
    public List<TreeNode> helper(int start, int end) {
        List<TreeNode> res = new ArrayList<TreeNode>();
        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 rootLeft: left) {
                for (TreeNode rootRight: right) {
                    TreeNode root = new TreeNode(i);
                    root.left = copy(rootLeft);
                    root.right = copy(rootRight);
                    res.add(root);
                }
            }
        }
        return res;
    }
    public TreeNode copy(TreeNode node) {
        if (node == null) return null;
        TreeNode copyNode = new TreeNode(node.val);
        copyNode.left = copy(node.left);
        copyNode.right = copy(node.right);
        return copyNode;
    }
}

可以删除copy()，简化为：

public class Solution {
    public List<TreeNode> generateTrees(int n) {
        return helper(1, n);
    }
    public List<TreeNode> helper(int start, int end) {
        List<TreeNode> res = new ArrayList<TreeNode>();
        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 nodeLeft: left) {
                for (TreeNode nodeRight: right) {
                    TreeNode root = new TreeNode(i);
                    root.left = nodeLeft;
                    root.right = nodeRight;
                    res.add(root);
                }
            }
        }
        return res;
    }
}