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二叉树根为空

[英]Binary Tree root is null

I'm trying to create a binary search tree but it doesn't seem to be working.我正在尝试创建一个二叉搜索树,但它似乎不起作用。 I debugged it, and it says the root is null.我调试了它,它说根为空。 I don't understand why it is null.我不明白为什么它是空的。 I set it to null initially in the constructor, but then when I call the insert() method, it is no longer null, right?我最初在构造函数中将它设置为 null,但是当我调用 insert() 方法时,它不再为 null,对吗? Can someone help me understand this.有人可以帮助我理解这一点。 Thanks.谢谢。

#include "stdafx.h"
    #include <iostream>

using namespace std;


struct node
{
public:
    int value;
    node * left;
    node * right;
};


class bTree
{
public:
    node * root;

public:
    bTree();
    void insert(node * r, int val);
    void insert(int val);
    void traversePreorder();
    void traversePreorder(node * r);


};

bTree::bTree()
{
    root = NULL;
}

void bTree::insert(node * r, int val)
{
    if (r == NULL)
    {
        r = new node();
        r->value = val;
        r->left = NULL;
        r->right = NULL;
        return;
    }
    else
    {
        if (val <= r->value)
        {
            insert(r->left, val);
        }
        else
        {
            insert(r->right, val);
        }
    }
}

void bTree::insert(int val)
{
    insert(root, val);
}

void bTree::traversePreorder(node * r)
{
    if (root == nullptr)
        return;
    else
    {
        cout << root->value << " ";
        traversePreorder(root->left);
        traversePreorder(root->right);
    }
}

void bTree::traversePreorder()
{
    traversePreorder(root);
}

int main()
{
    bTree * myTree = new bTree();

    myTree->insert(30);

    myTree->insert(40);
    myTree->insert(20);
    myTree->insert(10);
    myTree->insert(50);


    myTree->traversePreorder();

    return 0;
}

If you debug into void bTree::insert(node * r, int val) , you'll find that root does not get changed at all.如果你调试到void bTree::insert(node * r, int val) ,你会发现root根本没有改变。

In void bTree::insert(node * r, int val) , r is passed by value, so the change of r inside the function ( new etc) has nothing to do with the outside variable ( root ).void bTree::insert(node * r, int val)r是按值传递的,因此函数内部r的变化( new等)与外部变量( root )无关。 You could change it to pass by reference:您可以将其更改为通过引用传递:

void bTree::insert(node *& r, int val)

See What's the difference between passing by reference vs. passing by value?请参阅按引用传递与按值传递之间的区别是什么?
How to pass objects to functions in C++?如何将对象传递给 C++ 中的函数?

BTW: In void bTree::traversePreorder(node * r) , you should use r , not root :顺便说一句:在void bTree::traversePreorder(node * r) ,您应该使用r ,而不是root

void bTree::traversePreorder(node * r)
{
    if (r == nullptr) 
        return;
    else
    {
        cout << r->value << " ";
        traversePreorder(r->left);
        traversePreorder(r->right);
    }
}

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