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| 1 | +/* |
| 2 | + * @lc app=leetcode id=95 lang=java |
| 3 | + * |
| 4 | + * [95] Unique Binary Search Trees II |
| 5 | + */ |
| 6 | + |
| 7 | +// @lc code=start |
| 8 | +/** |
| 9 | + * Definition for a binary tree node. |
| 10 | + * public class TreeNode { |
| 11 | + * int val; |
| 12 | + * TreeNode left; |
| 13 | + * TreeNode right; |
| 14 | + * TreeNode() {} |
| 15 | + * TreeNode(int val) { this.val = val; } |
| 16 | + * TreeNode(int val, TreeNode left, TreeNode right) { |
| 17 | + * this.val = val; |
| 18 | + * this.left = left; |
| 19 | + * this.right = right; |
| 20 | + * } |
| 21 | + * } |
| 22 | + */ |
| 23 | +/** |
| 24 | + * Definition for a binary tree node. |
| 25 | + * public class TreeNode { |
| 26 | + * int val; |
| 27 | + * TreeNode left; |
| 28 | + * TreeNode right; |
| 29 | + * TreeNode() {} |
| 30 | + * TreeNode(int val) { this.val = val; } |
| 31 | + * TreeNode(int val, TreeNode left, TreeNode right) { |
| 32 | + * this.val = val; |
| 33 | + * this.left = left; |
| 34 | + * this.right = right; |
| 35 | + * } |
| 36 | + * } |
| 37 | + */ |
| 38 | +class Solution { |
| 39 | + /* |
| 40 | + 分左右区间进行recursion |
| 41 | + time: As we have to generate all the possible trees and the total number of possible |
| 42 | + trees possible is a catalan number which is (4^n)/((n^(3/2))*sqrt(pi)), |
| 43 | + which makes the time of this algorithm to be O(4^n). |
| 44 | + */ |
| 45 | + public List<TreeNode> generateTrees(int n) { |
| 46 | + return helper(1, n); |
| 47 | + } |
| 48 | + |
| 49 | + private List<TreeNode> helper(int low, int high) { |
| 50 | + List<TreeNode> res = new ArrayList<>(); |
| 51 | + if (low > high) return res; |
| 52 | + for (int i = low; i <= high; ++i) { |
| 53 | + List<TreeNode> left = helper(low, i - 1); |
| 54 | + List<TreeNode> right = helper(i + 1, high); |
| 55 | + if (left.size() == 0 && right.size() == 0) { |
| 56 | + res.add(new TreeNode(i)); |
| 57 | + } else if (left.size() == 0) { |
| 58 | + for (TreeNode r : right) { |
| 59 | + TreeNode root = new TreeNode(i); |
| 60 | + root.right = r; |
| 61 | + res.add(root); |
| 62 | + } |
| 63 | + } else if (right.size() == 0) { |
| 64 | + for (TreeNode l : left) { |
| 65 | + TreeNode root = new TreeNode(i); |
| 66 | + root.left = l; |
| 67 | + res.add(root); |
| 68 | + } |
| 69 | + } else { |
| 70 | + for (TreeNode l : left) { |
| 71 | + for (TreeNode r : right) { |
| 72 | + TreeNode root = new TreeNode(i); |
| 73 | + root.left = l; |
| 74 | + root.right = r; |
| 75 | + res.add(root); |
| 76 | + } |
| 77 | + } |
| 78 | + } |
| 79 | + } |
| 80 | + return res; |
| 81 | + } |
| 82 | +} |
| 83 | +// @lc code=end |
| 84 | + |
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