95. Unique Binary Search Trees II

Given an integern, generate all structurally uniqueBST's(binary search trees) that store values 1...n.

For example,
Givenn= 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

Analysis:

Divide and conquer approach. We pick a number i from 1 to n, call genTree(1, i-1), genTree(i+1, n) to generates all of its left tree and right tree's root. Then we do combination of these 2 list.

Complexity:

Time: O(2^n)

Space:

Code:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<TreeNode> generateTrees(int n) {
        if(n==0)
        {
            return new ArrayList<TreeNode>();
        }
        return genTree(1, n);
    }

    private List<TreeNode> genTree(int s, int e)
    {
        if(s > e)
        {
            return new ArrayList<TreeNode>();
        }

        List<TreeNode> returnList = new ArrayList<TreeNode>();

        for(int i=s; i<=e; i++)
        {
            List<TreeNode> left = genTree(s, i-1);
            List<TreeNode> right = genTree(i+1, e);
            if(left.size() == 0 && right.size() == 0)
            {
                returnList.add(new TreeNode(i));
            }
            else if(left.size() == 0)
            {
                for(TreeNode itr : right)
                {
                    TreeNode root = new TreeNode(i);
                    root.right = itr;
                    returnList.add(root);
                }
            }
            else if(right.size() == 0)
            {
                for(TreeNode itr : left)
                {
                    TreeNode root = new TreeNode(i);
                    root.left = itr;
                    returnList.add(root);
                }
            }
            else
            {
                for(TreeNode lnode : left)
                {
                    for(TreeNode rnode : right)
                    {
                        TreeNode root = new TreeNode(i);
                        root.left = lnode;
                        root.right = rnode;
                        returnList.add(root);
                    }
                }
            }
        }
        return returnList;
    }
}