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Unique Binary Search Trees II

时间:2016-07-12 21:17:59      阅读:173      评论:0      收藏:0      [点我收藏+]

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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
 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public List<TreeNode> generateTrees(int n) {
12         if (n == 0) {
13             return new ArrayList<TreeNode>();
14         }
15         return generate(1, n);
16     }
17     
18     private List<TreeNode> generate(int start, int end) {
19         List<TreeNode> result = new ArrayList<>();
20         if (start > end) {
21             result.add(null);
22             return result;
23         }
24         for (int i = start; i <= end; i++) {
25             List<TreeNode> left = generate(start, i - 1);
26             List<TreeNode> right = generate(i + 1, end);
27             for (TreeNode l : left) {
28                 for (TreeNode r : right) {
29                     TreeNode root = new TreeNode(i);
30                     root.left = l;
31                     root.right = r;
32                     result.add(root);
33                 }
34             }
35         }
36         return result;
37     }
38 }

 

Unique Binary Search Trees II

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原文地址:http://www.cnblogs.com/FLAGyuri/p/5664688.html

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