Assign the root value as head node value in linked list.
Do level-order traversal For each tree node create a single linked list node and append it to the linked list.
Display the single linked list.
How the tree is converted to the single list...
Let’s do solve the above example.
Root=2;
Queue status: 2
----------------------------------------------------
1st iteration
Queue not empty
Queue front is 2
Head is null, thus head is 2
Linked list up to now:2->NULL
Pop 2
Push: 2->left(7) & 2->right(5)
Queue status: 7, 5
----------------------------------------------------
2nd iteration
Queue not empty
Queue front is 7
Head is not null, thus append 7
Linked list up to now:2->7->NULL
Pop 7
Push: 7->left(2)& 7->right(6)
Queue status: 5, 2, 6
----------------------------------------------------
3rd iteration
Queue not empty
Queue front is 5
Head is not null, thus append 5
Linked list up to now:2->7->5->NULL
Pop 5
Push: 5->right (9) only (5->left is NULL)
Queue status: 2, 6, 9
----------------------------------------------------
4th iteration
Queue not empty
Queue front is 2
Head is not null, thus append 2
Linked list up to now:2->7->5->2->NULL
Pop 2
Push: Nothing ( both child are NULL)
Queue status: 6, 9
----------------------------------------------------
5th iteration
Queue not empty
Queue front is 6
Head is not null, thus append 6
Linked list up to now:2->7->5->2->6->NULL
Pop 6
Push: 6->left(5) and 6->right(11)
Queue status: 9, 5, 11
----------------------------------------------------
6th iteration
Queue not empty
Queue front is 9
Head is not null, thus append 9
Linked list up to now:2->7->5->2->6->9->NULL
Pop 9
Push: 9->left(4) only (right child NULL)
Queue status: 5, 11, 4
----------------------------------------------------
7th iteration
Queue not empty
Queue front is 5
Head is not null, thus append 5
Linked list up to now:2->7->5->2->6->9->5->NULL
Pop 5
Push: Nothing (both child NULL)
Queue status: 11, 4
----------------------------------------------------
8th iteration
Queue not empty
Queue front is 11
Head is not null, thus append 11
Linked list up to now:2->7->5->2->6->9->5->11->NULL
Pop 11
Push: Nothing (both child NULL)
Queue status: 4
----------------------------------------------------
8th iteration
Queue not empty
Queue front is 4
Head is not null, thus append 4
Linked list up to now: 2->7->5->2->6->9->5->11->4->NULL
Pop 4
Push: Nothing (both child NULL)
Queue status: empty queue
----------------------------------------------------
Iteration stopsSo final link list is: 2->7->5->2->6->9->5->11->4->NULL
C implementation to to convert a Binary Tree into a Singly Linked List by Traversing Level by Level
#include <bits/stdc++.h>
using namespace std;
// tree node is defined
class tree{
public:
int data;
tree *left;
tree *right;
};
class sll{
public:
int data;
sll* next;
};
sll* creatnode(int d){ //create node for single linked list
sll* temp=(sll*)malloc(sizeof(sll));
temp->data=d;
temp->next=NULL;
return temp;
}
void display(sll* head){
sll* current=head; // current node set to head
printf("displayig the converted list...\n");
while(current!=NULL){ //traverse until current node isn't NULL
if(current->next)
printf("%d->",current->data);
else
printf("%d->NULL\n",current->data);
current=current->next; // go to next node
}
}
sll* flatten(tree* root)
{
//Declare queue using STL
sll* head=NULL,*tempL;
queue<tree*> q;
//enqueue the root
q.push(root);
vector<int> store;
tree* temp;
//do the level order traversal & build single linked list
while(!q.empty()){
//dequeue
temp=q.front();
q.pop();
if(head==NULL){//for inserting first node
head=creatnode(temp->data);
tempL=head;
}
else{//for inserting rest of the nodes
tempL->next=creatnode(temp->data);
tempL=tempL->next;
}
// do level order traversing
if(temp->left)//for left child
q.push(temp->left);
if(temp->right)//for right child
q.push(temp->right);
}
return head;
}
tree* newnode(int data) // creating new node for tree
{
tree* node = (tree*)malloc(sizeof(tree));
node->data = data;
node->left = NULL;
node->right = NULL;
return(node);
}
int main()
{
//**same tree is builted as shown in example**
cout<<"same tree is built as shown in example\n";
tree *root=newnode(2);
root->left= newnode(7);
root->right= newnode(5);
root->right->right=newnode(9);
root->right->right->left=newnode(4);
root->left->left=newnode(2);
root->left->right=newnode(6);
root->left->right->left=newnode(5);
root->left->right->right=newnode(11);
cout<<"converting the tree into a single link list...\n";
cout<<"by traversing the tree level-wise\n";
sll* head=flatten(root);
//displaying the list built from the tree
display(head);
return 0;
}
Output
same tree is built as shown in example
converting the tree into a single link list...
by traversing the tree level-wise
displayig the converted list...
2->7->5->2->6->9->5->11->4->NULL
Algorithm:
For each tree node create a single linked list node and append it to the linked list.
How the tree is converted to the single list...
Let’s do solve the above example.
C implementation to to convert a Binary Tree into a Singly Linked List by Traversing Level by Level
Output