95. Unique Binary Search Trees II

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problem

solution

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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 l, int r){
        if(l>r) return {nullptr};
        vector<TreeNode *> ret;
        for(int i=l;i<=r;++i){
            vector<TreeNode*> left = generateTrees(l, i-1);
            vector<TreeNode*> right = generateTrees(i+1, r);
            for(auto l:left){
                for(auto r:right){
                    TreeNode *t = new TreeNode(i);
                    t->left = l;
                    t->right = r;
                    ret.push_back(t);
                }
            }
            
        }
        return ret;
        
    }
    vector<TreeNode*> generateTrees(int n) {
        return generateTrees(1, n);      
    }
};
  • memo pattern
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    class Solution {
    public:
        vector<vector<vector<TreeNode*>>> memo;
        vector<TreeNode*> generateTrees(int l, int r){
            if(l>r) return {nullptr};
            if(!memo[l][r].empty()) return memo[l][r];
            vector<TreeNode *> ret;
            for(int i=l;i<=r;++i){
                vector<TreeNode*> left = generateTrees(l, i-1);
                vector<TreeNode*> right = generateTrees(i+1, r);
                for(auto l:left){
                    for(auto r:right){
                        TreeNode *t = new TreeNode(i);
                        t->left = l;
                        t->right = r;
                        ret.push_back(t);
                    }
                }
                
            }
            return  memo[l][r] = ret;
            
        }
        vector<TreeNode*> generateTrees(int n) {
            memo = vector<vector<vector<TreeNode * >>>(n+1, vector<vector<TreeNode*>>(n+1));
            return generateTrees(1, n);
            
            
        }
    };

    analysis

  • time complexity O(n^2)