[LeetCode] Unique Binary Search Trees II

最新推荐文章于 2020-12-09 16:51:01 发布

转载最新推荐文章于 2020-12-09 16:51:01 发布 · 66 阅读

本文介绍了一种使用递归深度优先搜索的方法来生成所有可能的唯一二叉搜索树，并提供了详细的C++实现代码。对于给定的整数n，该算法能够找出所有结构上独一无二的二叉搜索树，这些树能存储从1到n的值。

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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.

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思路：Unique Binary Search Trees是用dp，求解所有的trees不能用dp。

对于每一个root，有leftNum * rightNum种情况，直接穷举，对于start>end, 返回含有NULL的一个节点，start==end也返回一个节点，框架和深搜一样

class Solution {
    public:
        vector<TreeNode *> dfs(int start, int end)
        {
            vector<TreeNode*> trees;

            if(start > end)
            {
                trees.push_back(NULL);
                return trees;
            }
            if(start == end)
            {
                TreeNode* p = new TreeNode(start);
                trees.push_back(p);
                return trees;
            }

            for(int rootVal = start; rootVal <= end; rootVal ++)
            {
                vector<TreeNode*> leftSubTrees = dfs(start, rootVal - 1);
                vector<TreeNode*> rightSubTrees = dfs(rootVal + 1, end);

                int leftSubTreesNum = leftSubTrees.size();
                int rightSubTreesNum = rightSubTrees.size();
                for(int i = 0; i < leftSubTreesNum; i++)
                {
                    for(int j= 0; j< rightSubTreesNum; j++)
                    {
                        TreeNode* root = new TreeNode(rootVal);
                        TreeNode* left = leftSubTrees[i];
                        TreeNode* right= rightSubTrees[j];
                        root->left = left;
                        root->right= right;
                        trees.push_back(root);
                    }
                }
            }
            return trees;
        }
        vector<TreeNode *> generateTrees(int n) {
            if(n == 0)
            {
                vector<TreeNode *>   trees;
                trees.push_back(NULL);
                return trees;
            }
            return dfs(1, n);
        }
};