LeetCode刷题笔录Unique Binary Search Trees II_given an integer n, return the number of structura-优快云博客

本文介绍了一种生成所有唯一二叉搜索树的方法，当给定一个整数n时，可以生成所有可能的不同结构的二叉搜索树，这些树将包含从1到n的值。通过递归地构建树的左子树和右子树，文章详细解释了如何使用循环遍历每个可能的根节点，并结合左右子树来构造最终的二叉树。

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

判断Unique的关键是：如果i是树的根，那么左子树由(1,i-1)组成，右子树由(i+1,n)组成.

用一个循环来遍历所有元素作为根，分别递归得到左右子树的List，然后把左右子树和根连接起来即可。

/**
 * 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 List<TreeNode> generateTrees(int n) {
        return generateTrees(1, n);
    }
    
    public List<TreeNode> generateTrees(int start, int end){
        List<TreeNode> result = new ArrayList<TreeNode>();
        if(start > end){
            result.add(null);
            return result;
        }
        
        for(int i = start; i <= end; i++){
            List<TreeNode> leftList = generateTrees(start, i - 1);
            List<TreeNode> rightList = generateTrees(i + 1, end);
            for(int j = 0; j < leftList.size(); j++){
                for(int k = 0; k < rightList.size(); k++){
                    TreeNode node = new TreeNode(i);
                    node.left = leftList.get(j);
                    node.right = rightList.get(k);
                    result.add(node);
                    
                }
            }
        }
        return result;
    }
}