95. Unique Binary Search Trees II

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95. Unique Binary Search Trees II
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Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1 ... n.

Example:

Input: 3
Output:
[
  [1,null,3,2],
  [3,2,null,1],
  [3,1,null,null,2],
  [2,1,3],
  [1,null,2,null,3]
]
Explanation:
The above output corresponds to the 5 unique BST's shown below:

   1         3     3      2      1
    \       /     /      / \      \
     3     2     1      1   3      2
    /     /       \                 \
   2     1         2                 3
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Difficulty:Medium
Total Accepted:114.1K
Total Submissions:342.3K
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Related Topics 
Dynamic ProgrammingTree
Similar Questions 
Unique Binary Search TreesDifferent Ways to Add Parentheses
Java	





right
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/**
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 * Definition for a binary tree node.
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 * public class TreeNode {
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 *     int val;
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 *     TreeNode left;
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 *     TreeNode right;
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 *     TreeNode(int x) { val = x; }
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 * }
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 */
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class Solution {
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     public List<TreeNode> generateTrees(int n ){
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         return generateTrees(1,n);
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     }
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    public List<TreeNode> generateTrees(int m, int n) {
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        if (m == n) {
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            ArrayList<TreeNode> treeNodes = new ArrayList<>();
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            treeNodes.add(new TreeNode(m));
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            return treeNodes;
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        }
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        List<TreeNode> list = new ArrayList<>();
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        for (int i = m; i <= n; i++) {
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            List<TreeNode> left = null;
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            List<TreeNode> right = null;
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            if (i == m) {
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                right = generateTrees(i + 1, n);
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            } else {
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                if (i == n) {
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                    left = generateTrees(m, i - 1);
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                } else {
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                    right = generateTrees(i + 1, n);
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                    left = generateTrees(m, i - 1);
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                }
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            }
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            if (left != null && right != null) {
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                for (TreeNode treeNode : left) {
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                    for (TreeNode node : right) {
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                        TreeNode node1 = new TreeNode(i);
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                        node1.left = treeNode;
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                        node1.right = node;
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                        list.add(node1);
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                    }
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                }
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            } else {
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                if (left != null) {
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                    for (TreeNode treeNode : left) {
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                        TreeNode node1 = new TreeNode(i);
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                        node1.left = treeNode;
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                        list.add(node1);
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​
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                    }
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                } else {
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                    for (TreeNode treeNode : right) {
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                        TreeNode node1 = new TreeNode(i);
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                        node1.right = treeNode;
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                        list.add(node1);
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                    }
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                }
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            }
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        }
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        return list;
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​
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    }
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​
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}
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