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);
	}

}