Perfect Binary Tree Specific Level Order Traversal - GeeksforGeeks

Perfect Binary Tree Specific Level Order Traversal - GeeksforGeeks
Given a Perfect Binary Tree like below:

Print the level order of nodes in following specific manner:

  1 2 3 4 7 5 6 8 15 9 14 10 13 11 12 16 31 17 30 18 29 19 28 20 27 21 26  22 25 23 24

i.e. print nodes in level order but nodes should be from left and right side alternatively. Here 1^st and 2^nd levels are trivial.
While 3^rd level: 4(left), 7(right), 5(left), 6(right) are printed.
While 4^th level: 8(left), 15(right), 9(left), 14(right), .. are printed.
While 5^th level: 16(left), 31(right), 17(left), 30(right), .. are printed.

Approach 2:
The standard level order traversal idea will slightly change here. Instead of processing ONE node at a time, we will process TWO nodes at a time. And while pushing children into queue, the enqueue order will be: 1^st node’s left child, 2^nd node’s right child, 1^st node’s right child and 2^nd node’s left child.

void printSpecificLevelOrder(Node *root)

{

    if (root == NULL)

        return;

 

    // Let us print root and next level first

    cout << root->data;

 

    // / Since it is perfect Binary Tree, right is not checked

    if (root->left != NULL)

      cout << " " << root->left->data << " " << root->right->data;

 

    // Do anything more if there are nodes at next level in

    // given perfect Binary Tree

    if (root->left->left == NULL)

        return;

 

    // Create a queue and enqueue left and right children of root

    queue <Node *> q;

    q.push(root->left);

    q.push(root->right);

 

    // We process two nodes at a time, so we need two variables

    // to store two front items of queue

    Node *first = NULL, *second = NULL;

 

    // traversal loop

    while (!q.empty())

    {

       // Pop two items from queue

       first = q.front();

       q.pop();

       second = q.front();

       q.pop();

 

       // Print children of first and second in reverse order

       cout << " " << first->left->data << " " << second->right->data;

       cout << " " << first->right->data << " " << second->left->data;

 

       // If first and second have grandchildren, enqueue them

       // in reverse order

       if (first->left->left != NULL)

       {

           q.push(first->left);

           q.push(second->right);

           q.push(first->right);

           q.push(second->left);

       }

    }

}

Followup Questions:

1. The above code prints specific level order from TOP to BOTTOM. How will you do specific level order traversal from BOTTOM to TOP (Amazon Interview | Set 120 – Round 1 Last Problem)
2. What if tree is not perfect, but complete.
3. What if tree is neither perfect, nor complete. It can be any general binary tree.

Approach 1:
We can do standard level order traversal here too but instead of printing nodes directly, we have to store nodes in current level in a temporary array or list 1^st and then take nodes from alternate ends (left and right) and print nodes. Keep repeating this for all levels.
This approach takes more memory than standard traversal.
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