leetcode 95: Unique Binary Search Trees II

最新推荐文章于 2022-01-13 04:43:51 发布

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本文介绍了一种通过递归方法生成所有可能的二叉搜索树(BST)的方法，并详细解释了如何保存这些BST。核心思想是在每一步递归中声明一个向量来存储左子树和右子树的所有组合。

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Learned from http://blog.youkuaiyun.com/lanxu_yy/article/details/17504837. The hard thing about this problem is not how to generate a BST, but how to save the BST when one is generated. The idea here is to declare the vector in every recursion step and save all its left and right subtrees. Let's say I got m left subtrees and n right subtrees, I then generate m*n roots and push all these roots into the vector and return it.

/**
 * Definition for a binary tree node.
 * 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) {
        return helper(1,n);
    }
    vector<TreeNode*> helper(int start,int end)
    {
        vector<TreeNode*> subtree;
        if(start>end)
        {
            subtree.push_back(NULL);
            return subtree;
        }
        for(int i=start;i<=end;i++)
        {
            vector<TreeNode*> left=helper(start,i-1);
            vector<TreeNode*> right=helper(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];
                    subtree.push_back(root);
                }
        }
        return subtree;
    }
};