【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
;
}
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