Printing all root to leaf paths - c++

Question

The code that I tried to print all root to leaf paths in a Binary Tree.
#include<iostream>
#include<stack>
using namespace std;
bool visited[100];
void intilize(){
for(int i=0;i<100;i++)
visited[i]=false;
}
struct node
{
int data;
struct node *left,*right;
};
struct node* createNode(int k)
{
struct node* temp = new node;
temp->left = NULL;
temp->right = NULL;
temp->data = k;
return temp;
}
stack<node*> s,s1;
void print(){
while(!s.empty()){
s1.push(s.top());
s.pop();
}
while(!s1.empty()){
struct node* a= s1.top();
cout<<a->data<<" ";
s1.pop();
s.push(a);
if(s1.empty())
return;
}
}
void printpath(struct node* node){
if(node==NULL) return;
s.push(node);
while(!s.empty()){
struct node* top=s.top();
visited[top->data]=true;
if(top->left!=NULL&&visited[top->left->data]==false)
printpath(top->left);
else if(top->right!=NULL&&visited[top->right->data]==false)
printpath(top->right);
else if(top->left==NULL&&top->right==NULL){
print();
cout<<"\n";
}
s.pop();
}
}
int main() {
struct node* root = createNode(50);
root->left = createNode(7);
root->right = createNode(2);
root->right->left = createNode(1);
root->right->right = createNode(30);
root->right->right->right = createNode(40);
root->right->left->left = createNode(10);
root->right->left->left->left = createNode(12);
intilize();
printpath(root);
return 0;
}
The code gives segmentation fault because there is some issue with the termination condition.
Can someone help me in figuring out the problem.

Answer 1

That approach is over-complicated, and fragile.
A separate stack is not needed for this.
The separate "visible" array is not needed for this.
All that's needed is a stock recursive visitor that recursively descends into this tree, that also takes an additional parameter to a structure that's dynamically built on the stack, which builds the path to the root on the fly, using a link list that goes something like this:
struct path_to_root {
struct path_to_root *next;
struct node *n;
};
Now, all that's needed to print a path to each leaf note is a bog-standard visitor, that recursively iterates over the tree, and this additional parameter. Here's a rough idea of the general approach:
void printpath(struct node *n, struct path_to_root *p)
{
struct path_to_root pnext;
if (!n)
return;
if (!n->left && !n->right)
{
/* Your homework assignment here is to print the path that's in "p" */
}
pnext.n=n;
pnext.next=p;
printpath(n->left, &pnext);
printpath(n->right, &pnext);
}
And this would be invoked as:
printpath(root, NULL);
Your homework assignment, as noted, is to implement the actual code that prints the path to the leaf, using the p parameter, in the space indicated. At that point, the path to the leaf will be found in the p parameter.
Now, one tricky part here is that p will be the leaf's parent, p->next will be its grandparent, and so on. So the path is from the bottom to the top, not top to bottom, but that's a minor detail, that can be handled in the print code.
Or, alternatively, it wouldn't be too much extra work to dynamically build the path to the leaf from top to bottom, in the same manner.