leetcode------Unique Binary Search Trees II

标题: Unique Binary Search Trees II
通过率: 27.5%
难度: 中等

Given 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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

在第一个版本中要求的是说出BST数量,本题要求把BST全部都输出,用list去储存一个每个可解方案的root,跟第一版本的类型相似,尝试每一个元素作为root那么就把一组数分成了左右,然后去处理左边和右边,这样就把全部类型都列举出来了,还是递归的思想;

具体看代码:

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; left = null; right = null; }
 8  * }
 9  */
10 public class Solution {
11     public List<TreeNode> generateTrees(int n) {
12         return getTree(1,n);
13     }
14     public ArrayList<TreeNode> getTree(int left,int right){
15         ArrayList<TreeNode> result=new ArrayList<TreeNode>();
16         if(left>right){
17             result.add(null);
18             return result;
19         }
20         for(int i=left;i<=right;i++){
21             ArrayList<TreeNode> leftt=getTree(left,i-1);
22             ArrayList<TreeNode> rightt=getTree(i+1,right);
23             for(int j=0;j<leftt.size();j++){
24                 for(int k=0;k<rightt.size();k++){
25                     TreeNode root=new TreeNode(i);
26                     root.left=leftt.get(j);
27                     root.right=rightt.get(k);
28                     result.add(root);
29                 }
30             }
31         }
32         return result;
33     }
34 }
原文地址:https://www.cnblogs.com/pkuYang/p/4296499.html