Level Order Traversal: Deleting a Subtree - c++

Question

#include <iostream>
using namespace std;
struct node {
int item;
node* l;
node* r;
node (int x) {
item = x;
l = 0;
r = 0;
}
node(int x, node* l, node* r) {
item = x;
this->l = l;
this->r = r;
}
};
typedef node* link;
class QUEUE {
private:
link* q;
int N;
int head;
int tail;
public:
QUEUE(int maxN) {
q = new link[maxN + 1];
N = maxN + 1;
head = N;
tail = 0;
}
int empty() const {
return head % N == tail;
}
void put(link item) {
q[tail++] = item;
tail = tail % N;
}
link get() {
head = head % N;
return q[head++];
}
};
link head = 0; // Initial head of the tree
link find(int x) {
if (head == 0) {
cout << "\nEmpty Tree\n";
return 0;
}
link temp = head;
// To find the node with the value x and return its link
QUEUE q(100);
q.put(temp);
while (!q.empty()) {
temp = q.get();
if (temp->item == x) {
return temp;
}
if (temp->l != 0) q.put(temp->l);
if (temp->r != 0) q.put(temp->r);
}
return 0;
}
void print(link temp) {
QUEUE q(100);
q.put(temp);
while (!q.empty()) {
temp = q.get();
cout << temp->item << ", ";
if (temp->l != 0) q.put(temp->l);
if (temp->r != 0) q.put(temp->r);
}
}
void deleteAll(link h) {
// This deletes the entire binary tree
QUEUE q(100);
q.put(h);
while (!q.empty()) {
h = q.get();
if (h->l != 0) q.put(h->l);
if (h->r != 0) q.put(h->r);
delete h;
}
}
int main() {
link temp = 0;
char c;
int n1, n2;
cout << "\n\nPlease enter the input instructions (X to exit program) : \n\n";
do {
cin >> c;
switch (c) {
case 'C': cin >> n1;
if (head == 0) {
head = new node(n1);
cout << "\nRoot node with item " << n1 << " has been created\n\n";
} else {
cout << "\nError: Tree is not empty\n\n";
}
break;
case 'L': cin >> n1 >> n2;
temp = find(n1);
if (temp != 0) {
if (temp->l == 0) {
temp->l = new node(n2);
cout << "\nNode with item " << n2 << " has been added\n\n";
}
else {
cout << "\nError: The specified node already has a left child\n\n";
}
}
else {
cout << "\nError: The specified node doesn't exist\n\n";
}
break;
case 'R': cin >> n1 >> n2;
temp = find(n1);
if (temp != 0) {
if (temp->r == 0) {
temp->r = new node(n2);
cout << "\nNode with item " << n2 << " has been added\n\n";
}
else {
cout << "\nError: The specified node already has a right child\n\n";
}
}
else {
cout << "\nError: The specified node doesn't exist\n\n";
}
break;
case 'P': cin >> n1;
temp = find(n1);
if (head != 0) {
cout << "\nLevel-order traversal of the entire tree: ";
print(temp);
}
else {
cout << "\nError: No elements to print\n\n";
}
break;
case 'D': cin >> n1;
temp = find(n1);
deleteAll(temp);
temp = 0;
break;
case 'X': cout << "\nExiting Program\n\n";
break;
default: cout << "\nInvalid input entered. Try again.\n\n";
}
} while (c != 'X');
system("pause");
return 0;
}
Sample Input:
C 9
L 9 8
R 9 6
L 8 3
R 8 5
R 6 2
L 3 4
L 4 10
L 5 1
R 5 11
L 1 12
R 1 7
It all works fine until I delete a subtree and print when it prints garbage value before crashing. Please help me figure out the bug because I've been trying in vain for hours now.
It all works fine until I delete a subtree and print when it prints garbage value before crashing. Please help me figure out the bug because I've been trying in vain for hours now.

Answer 1

Try the recursive function:
void Delete(link h)
{
if(h)
{
if(h->l) Delete(h->l);
if(h->r) Delete(h->r);
delete(h);
}
}

Answer 2

When you delete a node, you call deleteAll(temp) which deletes temp, but it doesn't remove the pointer value from the l or r of temp's parent node.
This leaves you with a invalid pointer, causing garbage printing and crashing.
Unfortunately, the way your find works currently, you don't know what the current temp node's parent is when you get around to checking its value.
One way to fix it is to have a different type of find (called something like remove) that looks in l and r at each iteration for the value and sets l or r to NULL before returning the pointer. You might have to have a special case for when the value is found in the root.
Edit (sample code added):
I am assuming you are not using recursion for some reason, so my code uses your existing queue based code. I only changed enough to get it working.
findAndUnlink find the node with the value given and "unlinks" it from the tree. It returns the node found, giving you a completely separate tree. Note: it is up to the caller to free up the returned tree, otherwise you will leak memory.
This is a drop in replacement for find in your existing code, as your existing code then calls deleteAll on the returned node.
link findAndUnlink(int x) {
if (head == 0) {
cout << "\nEmpty Tree\n";
return 0;
}
link temp = head;
if (temp->item == x) {
// remove whole tree
head = NULL;
return temp;
}
// To find the node with the value x and remove it from the tree and return its link
QUEUE q(100);
q.put(temp);
while (!q.empty()) {
temp = q.get();
if (temp->l != NULL) {
if (temp->l->item == x) {
link returnLink = temp->l;
temp->l = NULL;
return returnLink;
}
q.put(temp->l);
}
if (temp->r != NULL) {
if (temp->r->item == x) {
link returnLink = temp->r;
temp->r = NULL;
return returnLink;
}
q.put(temp->r);
}
}
return 0;
}