LeetCode_Unique Binary Search Trees II

一.題目

Unique Binary Search Trees II

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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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

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二.解題技巧

    這道題和Unique Binary Search Trees很相似，只是這道題可能無法使用動態規劃減少計算的時間，因爲每一個節點的值是不一樣的，不存在相同的子問題的情況，因此，計算複雜度無法下降。使用遞歸的方式進行的時候，時間複雜度爲O(n^3)，空間複雜度爲O(n)。

三.實現代碼

#include <iostream>
#include <vector>
#include <unordered_map>

using std::vector;
using std::unordered_map;


/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/


struct TreeNode
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};


class Solution
{
private:
    unordered_map<int, vector<TreeNode*> > PreResult;

    vector<TreeNode*> generateTrees(int n, int base)
    {
        vector<TreeNode*> Result;

        if (n == 0)
        {
            TreeNode* TmpNode = NULL;
            Result.push_back(TmpNode);
            return Result;
        }

        if (n == 1)
        {
            TreeNode* TmpNode = new TreeNode(n + base);
            Result.push_back(TmpNode);
            return Result;
        }

        for (int HeadIndex = 1; HeadIndex <= n; HeadIndex++)
        {
            vector<TreeNode*> LeftChildVector = generateTrees(HeadIndex - 1, base);
            vector<TreeNode*> RightChildVector = generateTrees(n - HeadIndex, base + HeadIndex);

            const vector<TreeNode*>::size_type LEFTSIZE = LeftChildVector.size();
            const vector<TreeNode*>::size_type RIGHTSIZE = RightChildVector.size();

            for (vector<TreeNode*>::size_type IndexOfLeft = 0; IndexOfLeft < LEFTSIZE; IndexOfLeft++)
            {
                for (vector<TreeNode*>::size_type IndexOfRight = 0; IndexOfRight < RIGHTSIZE; IndexOfRight++)
                {
                    TreeNode *TmpHeadNode = new TreeNode(base + HeadIndex);
                    TmpHeadNode->left = LeftChildVector[IndexOfLeft];
                    TmpHeadNode->right = RightChildVector[IndexOfRight];

                    Result.push_back(TmpHeadNode);
                }
            }
        }

        return Result;

    }

public:
    vector<TreeNode*> generateTrees(int n)
    {
        return generateTrees(n, 0);
    }
};

四.體會

  這道題也是一個和Unique Binary Search Trees類似的問題，解題思路也比較類似，只是這個的返回是所有可能的二叉查找樹，主要解答過程只是稍微修改下即可。

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