Cinte's Leetcode Record
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95. Unique Binary Search Trees II

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Symbols count in article: 2k Reading time ≈ 2 mins.

95. Unique Binary Search Trees II

dp
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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<TreeNode*> generateTrees(int n) {
        vector<vector<TreeNode*>> dp(n + 1);
        dp[0] = {nullptr};
        for(size_t i = 1; i <= n; ++i)
        {
            for(size_t j = 1; j <= i; ++j)
            {
                for(auto& nodeLeft : dp[j - 1])
                {
                    for(auto& nodeRight : dp[i - j])
                    {
                        TreeNode* tmp = new TreeNode(j);
                        tmp->left = nodeLeft;
                        tmp->right = clone(nodeRight, j);
                        dp[i].push_back(tmp);
                    }
                }
            }
        }
        return dp.back();
    }
private:
    TreeNode* clone(TreeNode* root, int offset)
    {
        if(!root)
            return nullptr;
        TreeNode* newNode = new TreeNode(root->val + offset);
        newNode->left = clone(root->left, offset);
        newNode->right = clone(root->right, offset);
        return newNode;
    }
};

和Unique Binary Search Trees相似，只不过这个要列举出所有节点

递归分治
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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<TreeNode*> generateTrees(int n) {
        return helper(1, n);
    }
private:
    vector<TreeNode*> helper(int start, int end)
    {
        if(start > end)
            return { nullptr };
        vector<TreeNode*> ret;
        for(int i = start; i <= end; ++i)
        {
            auto left = helper(start, i - 1);
            auto right = helper(i + 1, end);
            for(auto& l : left)
            {
                for(auto& r : right)
                {
                    auto cur = new TreeNode(i);
                    cur->left = l;
                    cur->right = r;
                    ret.push_back(cur);
                }
            }
        }
        return ret;
    }
};

好难想

从下建立起树，然后往上走，连接