Unique Binary search Tree

Problem

Given an integer n, generate all structurally unique BST’s (binary search trees) that store values 1…n.

For 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

Solution 1

using recursion,

/**
 * 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) {
        List<TreeNode> result = new ArrayList<TreeNode>();
        if(n==0){
            return result;
        }
        return buildTrees(1,n);
    }
    public List<TreeNode> buildTrees(int start, int end){
      List<TreeNode> result = new ArrayList<TreeNode>();
      if(start>end ){
          result.add(null);
          return result;
      }
      List<TreeNode> rightTrees = new ArrayList<TreeNode>();
      List<TreeNode> leftTrees = new ArrayList<TreeNode>();
      // pick up a root
      for(int i = start; i <= end; i++){
          // since i is root, left tree would be from start -> i-1
          leftTrees = buildTrees(start, i-1);
          // right trees would be from i+1 to end
          rightTrees = buildTrees(i+1,end);
          // generate trees
          for(TreeNode leftTree : leftTrees){
            for(TreeNode rightTree: rightTrees){
              TreeNode tree = new TreeNode(i);
              tree.left = leftTree;
              tree.right = rightTree;
              result.add(tree);
          }
      }
      }
      return result;
    }
}

the result :

notice the above solution doesn’t use memorization

Solution 2 (with memorization)

/**
 * 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) {
        List<TreeNode> result = new ArrayList<TreeNode>();
        if(n==0){
            return result;
        }
        return buildTrees(1,n);
    }
    public List<TreeNode> buildTrees(int start, int end){
      List<TreeNode> result = new ArrayList<TreeNode>();
      Map<String,List<TreeNode>> hashMap = new HashMap<>();
      StringBuilder sb = new StringBuilder();
      sb.append(start).append("-").append(end);
      if(hashMap.get(sb.toString())!=null){
          return hashMap.get(sb.toString());
      }
      if(start>end ){
          result.add(null);
          return result;
      }
      List<TreeNode> rightTrees = new ArrayList<TreeNode>();
      List<TreeNode> leftTrees = new ArrayList<TreeNode>();
      // pick up a root
      for(int i = start; i <= end; i++){
          // since i is root, left tree would be from start -> i-1
          leftTrees = buildTrees(start, i-1);
          // right trees would be from i+1 to end
          rightTrees = buildTrees(i+1,end);
          // generate trees
          for(TreeNode leftTree : leftTrees){
            for(TreeNode rightTree: rightTrees){
              TreeNode tree = new TreeNode(i);
              tree.left = leftTree;
              tree.right = rightTree;
              result.add(tree);
          }
        }
      }
      hashMap.put(sb.toString(),result);
      return result;
    }
}

I use sart - end as key, List<TreeNode> as values to memorize the results

but the performance is even worse. I believe this is caused by the small number of test cases.

From WizNote