LeetCode | Unique Binary Search Trees II

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

/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; left = null; right = null; }
* }
*/

public class Solution {

public ArrayList<TreeNode> generateTree(int start, int end) {
ArrayList<TreeNode> result = new ArrayList<TreeNode>();
if (start > end) {
result.add(null);
return result;
}
ArrayList<TreeNode> leftTree = new ArrayList<TreeNode>();
ArrayList<TreeNode> rightTree = new ArrayList<TreeNode>();
for (int i = start; i <= end; i++) {                //从start到end，选取每一个数尝试构造BST
leftTree = generateTree(start, i-1);        //取小于i的数构造i的左子树
rightTree = generateTree(i+1, end);         //取大于i的数构造i的右子树，以此来保证i的二叉查找树有序性
for (int j = 0; j < leftTree.size(); j++) {
for (int k = 0; k < rightTree.size(); k++) {   //取i作为root构造BST，共有num_left*num_right种方式
TreeNode curNode = new TreeNode(i + 1);
curNode.left = leftTree.get(j);        //left、right是递归得到的root的集合，相当于list的每个节点都是一颗二叉查找树
curNode.right = rightTree.get(k);      //将每一种可能的组合，接到当前的root(i)上组成一颗BST
result.add(curNode);
}
}
}
return result;
}

//主调用函数：返回的List是每一棵树的root的集合，List.length = n的二叉搜索树的个数
public ArrayList<TreeNode> generateTrees(int n) {
return generateTree(0, n-1);
}

}