[Error]pointer being freed was not allocated

Question

I was practicing the code of BFS of tree with STL in c++ and I got one runtime error that i can't debug.everything is working fine if i dont call the printout() function . Please help as I am new to STL..

#include<iostream>
#include<malloc.h> //on llvm we don't need this
#include<list>
using namespace std;
typedef struct Node{
int val;
struct Node* left;
struct Node* right;
}node;
void push(node** root,int val)
{
    if(!(*root))
    {
        node* temp=(node*)malloc(sizeof(node));
        temp->val=val;
        temp->right=temp->left=NULL;
        *root=temp;
    }
    else if(val<(*root)->val)
        push(&((*root)->left),val);
    else
        push(&((*root)->right),val);
}

void printout(node* head)
{
    node* temp;
    temp=head;
    list<node*>qu;

    //using bfs here
    while(temp!=NULL)
    {
        cout<<temp->val<<endl;
        if(temp->left!=NULL)
            qu.push_back(temp->left);
        if(temp->right!=NULL)
            qu.push_back(temp->right);
        temp=qu.front();
        qu.pop_front();
        //free(temp);
    }
}

int main()
{
node* root=NULL;
push(&root,3);
push(&root,4);
push(&root,1);
push(&root,10);
push(&root,2);
printout(root);
}

though it is printing corrent output but with runtime time

3
1
4
2
10
a.out(613) malloc: *** error for object 0x7fff55ed8bc8: pointer being freed was not allocated
*** set a breakpoint in malloc_error_break to debug
Abort trap: 6

Answer 1

You are calling qu.front() in each iteration without checking if qu is empty. If it is empty - and finally it will be - your code breaks.

The simplest solution would be to check if qu is empty:

if (qu.empty()) {
    temp = NULL;
} else {
    temp=qu.front();
    qu.pop_front();
    //free(temp);
}

However, that looks weird. I would change the loop completely and use !qu.empty() as condition for the while loop.

list<node*> qu;
qu.push_back(head);
while(!qu.empty()) {
    node* temp = qu.front();
    qu.pop_front();
    if(temp->left)
        qu.push_back(temp->left);
    if(temp->right)
        qu.push_back(temp->right);
    //free(temp);
}

Answer 2

What happens is that when you get to the final "leaf" in the tree, the temp->left and temp->right are both NULL and you get an empty qu list. Calling qu.front() causes undefined behavior on an empty list: http://en.cppreference.com/w/cpp/container/list/front

You could add a size check before calling front.