Binary Tree root is null

Question

I'm trying to create a binary search tree but it doesn't seem to be working. I debugged it, and it says the root is null. I don't understand why it is null. I set it to null initially in the constructor, but then when I call the insert() method, it is no longer null, right? Can someone help me understand this. Thanks.

#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;
}

score 2 · Accepted Answer · edited May 23 '17 at 11:58

2

If you debug into void bTree::insert(node * r, int val), you'll find that root does not get changed at all.

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). You could change it to pass by reference:

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

See What's the difference between passing by reference vs. passing by value?
How to pass objects to functions in C++?

BTW: In void bTree::traversePreorder(node * r), you should use r, not root:

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

LIVE

edited May 23 '17 at 11:58

Community

1
1

answered Sep 10 '15 at 02:39

songyuanyao

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Can you explain another thing, I know I should ask a separate question but it's very on topic. Why does the addNode() function work in my linked list implementation seen in this link [link](http://ideone.com/7v5TXe) because that one is passing a node pointer and is able to manipulate that without a reference? Just a bit confused. Thanks! – smokenstein Sep 10 '15 at 03:09
@smokenstein Yes. It's passed by pointer, and you're changing the content the pointer points to, (not the pointer itself,) it's valid. – songyuanyao Sep 10 '15 at 03:19

Binary Tree root is null

1 Answers1

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