unique-binary-search-trees-ii

最新推荐文章于 2018-08-21 10:10:42 发布

llwjason5555

最新推荐文章于 2018-08-21 10:10:42 发布

阅读量247

点赞数

分类专栏： leetcode 文章标签： leetcode 递归

本文链接：https://blog.youkuaiyun.com/llwjason5555/article/details/78088511

版权

leetcode 专栏收录该内容

116 篇文章

订阅专栏

本文介绍了一个算法问题，即如何生成所有可能的唯一二叉搜索树（BST），这些树可以存储从1到n的值。通过递归方法实现，示例中展示了当n等于3时的5种不同BST结构。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目：

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.

OJ’s Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where ‘#’ signifies a path terminator where no node exists below.
Here’s an example:
1
/ \
2 3
/
4
\
5
The above binary tree is serialized as”{1,2,3,#,#,4,#,#,5}”.

程序：

class Solution {
public:
    vector<TreeNode *> generateTrees(int n) {
        if (n <= 0) return helper(1, 0);
        return helper(1,n);
    }
private:
    vector<TreeNode*> helper(int start,int end)
    {
        vector<TreeNode*> subTree;
        if(start>end)
        {
            subTree.push_back(nullptr);
            return subTree;
        }
        for (int k = start; k <= end; k++)
        {
           vector<TreeNode*> leftSubs = helper(start, k - 1);
           vector<TreeNode*> rightSubs = helper(k + 1, end);
           for (auto i : leftSubs)
           {
              for (auto j : rightSubs)
              {
                TreeNode *node = new TreeNode(k);
                node->left = i;
                node->right = j;
                subTree.push_back(node);
              }
            }
        }
        return subTree;
         
    }      
};