【LeetCode】95.Unique Binary Search Trees II(Medium)解题报告

【LeetCode】95.Unique Binary Search Trees II(Medium)解题报告

题目地址：https://leetcode.com/problems/unique-binary-search-trees-ii/description/
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

Solution1:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 96基础上，要把树构建出来，并存储
 两种方法，第一种dp，第二种dfs根绝二叉搜索树性质做，但是第二种方法更好
 第一种：DP
 */
class Solution {
    public List<TreeNode> generateTrees(int n) {
        List<TreeNode>[] res = new List[n+1];
        res[0] = new ArrayList<>();
        if(n == 0) return res[0];
        res[0].add(null);
        for(int i=1 ; i<=n ; i++){
            res[i] = new ArrayList<>();
            for(int j=0 ; j<i ; j++){
                for(TreeNode left : res[j]){
                    for(TreeNode right : res[i-j-1]){
                        TreeNode root = new TreeNode(j+1);
                        root.left = left;
                        root.right = clone(right,j+1);
                        res[i].add(root);
                    }
                }
            }
        }
        return res[n];
    }
    public TreeNode clone(TreeNode root,int k){
        if(root == null) return root;
        TreeNode cur = new TreeNode(root.val+k);
        cur.left = clone(root.left,k);
        cur.right = clone(root.right,k);
        return cur;
    }
}

Solution2:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 96基础上，要把树构建出来，并存储
 两种方法，第一种dp，第二种dfs根绝二叉搜索树性质做，但是第二种方法更好
 第二种
 */
class Solution {
    public List<TreeNode> generateTrees(int n) {
        if(n == 0) return new ArrayList<>();
        return genTreeList(1,n);
    }
    public List<TreeNode> genTreeList(int start,int end){
        List<TreeNode> list = new ArrayList<>();
        if(start>end){
            list.add(null);
        }
        for(int idx=start ; idx<=end ; idx++){
           List<TreeNode> leftList = genTreeList(start , idx-1);
            List<TreeNode> rightList = genTreeList(idx+1,end);
            for(TreeNode left : leftList){
                for(TreeNode right : rightList){
                    TreeNode root = new TreeNode(idx);
                    root.left = left;
                    root.right = right;
                    list.add(root);
                }
            } 
        }
        return list;
    }
}

Date：2018年3月29日