Unique Binary Search Trees II

最新推荐文章于 2021-09-11 12:08:06 发布

zephyr_be_brave

最新推荐文章于 2021-09-11 12:08:06 发布

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分类专栏： leetcode习题分析文章标签：二叉树构造

本文链接：https://blog.youkuaiyun.com/zephyr_be_brave/article/details/10853571

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本文介绍了一种生成从1到n的所有结构独特二叉搜索树的高效算法，并通过优化减少对象拷贝，显著提高性能。

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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

解答：划分左右子树分别构造。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<TreeNode *> generateTrees(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<TreeNode *> ret = makeTree(1,n);
        return ret;
    }
    
    vector<TreeNode*> makeTree(int start,int end){
        vector<TreeNode*> ret;
        if(start > end){
            ret.push_back(NULL);
            return ret;
        }
        if(start == end){
            TreeNode* root = new TreeNode(start);
            ret.push_back(root);
            return ret;
        }
        
        for(int i = start;i <= end;i++)
        {
            vector<TreeNode*> left = makeTree(start,i-1);
            vector<TreeNode*> right = makeTree(i+1,end);
            
            for(int j = 0;j < left.size();j++)
            {  
                for(int k = 0;k < right.size();k++)
                {
                    TreeNode* root = new TreeNode(i);
                    root->left = left[j];
                    root->right = right[k];
                    ret.push_back(root);
                }
            }
        }
        return ret;
    }
};

116 milli secs.

写完看到了一篇分析，说上面的function存在大量的对象拷贝，因为所有变量都是在栈上开辟，所以返回值的时候都需要通过拷贝构造函数来重构vector。可以把代码改成下面的样子解决这个问题。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<TreeNode *> generateTrees(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<TreeNode *>* ret = makeTree(1,n);
        return *ret;
    }
    
    vector<TreeNode*>* makeTree(int start,int end){
        vector<TreeNode*>* ret = new vector<TreeNode*>();
        if(start > end){
            ret->push_back(NULL);
            return ret;
        }
        if(start == end){
            TreeNode* root = new TreeNode(start);
            ret->push_back(root);
            return ret;
        }
        
        for(int i = start;i <= end;i++)
        {
            vector<TreeNode*>* left = makeTree(start,i-1);
            vector<TreeNode*>* right = makeTree(i+1,end);
            
            for(int j = 0;j < left->size();j++)
            {  
                for(int k = 0;k < right->size();k++)
                {
                    TreeNode* root = new TreeNode(i);
                    root->left = (*left)[j];
                    root->right = (*right)[k];
                    ret->push_back(root);
                }
            }
        }
        return ret;
    }
};

116 milli secs.