Question

我正在为C ++中的Tic-Tac-Toe创建一些基本的AI。它涉及创建一个由“板”类组成的二叉树。每个节点随机放置一个X或O，然后再创建两个子节点，直到有人赢了或者已经绘制了游戏。为了跟踪每个节点的值，我使用一个整数：-1表示O获胜，0表示抽奖，1表示X获胜，2表示游戏仍在进行中。 Here is an example of what a game could look like.

填充完树后，我需要查看根节点的左侧和右侧，并总结所有叶子。具有较高金额的一方将是最初两个孩子更理想的选择。

我有一些非常粗略的代码用于扩展看似有效的树，但是，当我尝试在扩展后总结我的树时，我似乎已经失去了每个节点包含的值。这是节点

的.cpp

BTNode.cpp

#include <iostream>
#include <string>
#include "BTNode.h"
#include <ctime>
#include <cstdlib>
#include <iostream>
using namespace std;

BTNode::BTNode(board gameboard, string setplayer)
    {
        value = 2; //assume the game is continuing
        cout << "Beginning a new node\n";
        nodeboard = gameboard;
        player = setplayer;
        expand();
    }

void BTNode::expand()
{
        //place a piece 
        cout << "Expanding";
        placerandom(player);
        //check to see if a leaf
        value = nodeboard.checkwinner(); //returns -1 for O win, 0 for draw, 1 for X win and 2 for game still in progress
        cout << "Value of this node: " << value << "\n";
        if (value == 2)
        {
            if (player == "X")
            {
                BTNode right(nodeboard, "O");
                BTNode left(nodeboard, "O");
            }
            else
            {
                BTNode left(nodeboard, "X");
                BTNode right(nodeboard, "X");
            }
        }
        else
        {
            cout << "A game had ended";
        }
}   

int BTNode::getvalue(int side)
        {
        //in case this node is the root, i will only want to check 1 side of the tree
        //-1 is left side, 0 is both, 1 is right
        int toreturn=0;
        if (value ==2)
            {
            if (side == -1)
                {
                toreturn = left ->getvalue(0);
                }
            if (side == 0)
                {
                toreturn = right->getvalue(0) + left->getvalue(0);
                }
            if (side == 1)
                {
                toreturn = right->getvalue(0);
                }
            }
            else
            {
            cout << "\nThis is a leaf, returning: " << value;
            toreturn = value;
            }
        return toreturn;
        }

void BTNode::placerandom(string player)
    {
    srand(time(NULL)); //make a new seed based off time
    int randx = rand()%3;
    int randy = rand()%3;
    while (not nodeboard.place(randx,randy,player))
        {
        randx = rand()%3;
        randy = rand()%3;
        }
    nodeboard.printboard();
    }

BTNode.h

#include <string>
#include "board.h"
using namespace std;
class BTNode
{
    public:
        BTNode(board gameboard, string setplayer);
        void placerandom(string player);
        int getvalue(int side);
        void expand();
    private:
        board nodeboard;
        string player;
        BTNode *left;
        BTNode *right;
        int value;
};

理论上，我应该能够从树中调用“root - ＆gt; getvalue（-1）”，这将返回从树左侧开始的所有节点的总和。但是，尝试在扩展函数之后从任何节点获取“值”返回“-1131609994”。

我创建了错误的树吗？在我试图总结之前，这似乎很有希望。

将数据存储在二叉树的节点中

0 个答案: