[Leetcode] Unique BST II

最新推荐文章于 2020-03-25 22:52:12 发布

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本文介绍了一种递归算法，用于生成所有可能的不同结构的二叉搜索树（BST），这些BST存储从1到n的值。以n=3为例，展示了如何通过递归地构建左子树和右子树来生成所有5种不同的BST结构。

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

递归，straightforward。

/**
 * 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
        return generateTree(1,n);
        
    }
    
    vector<TreeNode*> generateTree(int begin,int end)
    {
        
        vector<TreeNode*> result;
        
        if(begin>end)
        {
            result.push_back(NULL);
            return result;
        }
        
        for(int i = begin;i<=end;i++)
        {
            vector<TreeNode*> leftTree = generateTree(begin,i-1);
            vector<TreeNode*> rightTree = generateTree(i+1,end);
            for(int j = 0;j<leftTree.size();j++)
            {
                for(int k =0;k<rightTree.size();k++)
                {
                    TreeNode* current = new TreeNode(i);
                    current->left = leftTree[j];
                    current->right = rightTree[k];
                    result.push_back(current);
                }
            }
        }
    }
    
};