LeetCode-95. Unique Binary Search Trees II

题目描述

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

解题思路

利用分治算法的思想，不断将问题划分为子问题去求解，有一个问题就是关于节点克隆的问题，如果在每出现一种情况的时候都能新建root节点是可以避免掉这个问题的。

代码

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */class Solution {    //List<TreeNode> result = new LinkedList<TreeNode>();    public List<TreeNode> generateTrees(int n) {        if(n<=0)            return new LinkedList<TreeNode>();        return generateCore(1,n);//         List<TreeNode>[] temResult = new List[n+1];//         temResult[0] = new LinkedList<>();//         temResult[0].add(null);//         temResult[1]=new LinkedList<>();//         temResult[1].add(new TreeNode(1));//         for(int i=2;i<=n;i++){//             List<TreeNode> temList = new LinkedList<TreeNode>();//             for(int j=1;j<=i;j++){//                 for(TreeNode left:temResult[j-1]){//                     for(TreeNode right:generateCore(j+1,i)){//                         TreeNode root = new TreeNode(j);//                         root.left = left;//                         root.right = right;//                         temList.add(root);//                     }//                 }//             }//             temResult[i]=temList;//         }//         return temResult[n];    }    private List<TreeNode> generateCore(int start,int end){        List<TreeNode> tem = new LinkedList<TreeNode>();         if(start > end){           tem.add(null);            return tem;        }        for(int i=start;i<=end;i++){            List<TreeNode> leftList = generateCore(start,i-1);            List<TreeNode> rightList = generateCore(i+1,end);            for(TreeNode left:leftList){                for(TreeNode right:rightList){                    //这里每一种情况都创建一个新的root节点，避免后续操作覆盖                    TreeNode root = new TreeNode(i);                    root.left = left;                    root.right = right;                    tem.add(root);                }            }        }        return tem;    }}