Given

*n*, generate all structurally unique**BST's**(binary search trees) that store values 1...*n*.
For example,

Given

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

confused what

`"{1,#,2,3}"`

means? > read more on how binary tree is serialized on OJ.**Understand the problem:**

The main difference between the BST I is it requires to output all unique BSTs. So we can use recursion solution.

**Solution:**

/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; left = null; right = null; } * } */ public class Solution { public List<TreeNode> generateTrees(int n) { return generateTreesHelper(1, n); } private List<TreeNode> generateTreesHelper(int start, int end) { List<TreeNode> result = new ArrayList<TreeNode>(); if (start > end) { result.add(null); return result; } for (int i = start; i <= end; i++) { List<TreeNode> left = generateTreesHelper(start, i - 1); List<TreeNode> right = generateTreesHelper(i + 1, end); for (TreeNode l : left) { for (TreeNode r : right) { TreeNode root = new TreeNode(i); root.left = l; root.right = r; result.add(root); } } } return result; } }

this answer is wrong

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