Binary tree and especial node printing - c++

Question

I have a tree like this in binary representation;
1
/
2
\
3
/ \
6 4
\ \
7 5
/
8
/
9
\
10
\
11
But in reality this is not binary tree but like
1
/ | | \
2 3 4 5
/\ |
6 7 8
/| \
9 10 11
Can you please help me getting printed out something like(childs are printed out in reversed order)
1 : 5 4 3 2
5 : 8
3 : 7 6
8 : 11 10 9
My TNode class looks like.
class TNode {
public:
unsigned long int data;
TNode * left;///first child
TNode * right;/// sibling
TNode * parent;/// parent
TNode(unsigned long int d = 0, TNode * p = NULL, TNode * n = NULL)///konstruktors
{
left = NULL;
right = NULL;
parent = p;
data = d;
}
};
Does this need stack implementation?

Answer 1

What about this approach:
Traverse the tree in preorder (visiting each node). For each node (when travelled to it) check if it has a left child. If so (indicating a node with child elements in original tree) print the nodes data with a ':' and take its subtree and follow all right sons (which are representing all siblings) recursively, afterwards print each right sons data (you have it in reversed order.
In Code:
void print_siblings() {
if (this->right != NULL) {
this->right->print_siblings();
}
cout << data << ", ";
}
void traverse(void) {
if (this->left != NULL) {
cout << data << ":";
this->left->print_siblings();
cout << endl;
this->left->traverse();
}
if (this->right != NULL) {
this->right->traverse();
}
}
Edit: Here is a inorder-traversal:
void traverse_inorder(void) {
if (this->left != NULL) {
this->left->traverse_inorder();
cout << data << ":";
this->left->print_siblings();
cout << endl;
}
if (this->right != NULL) {
this->right->traverse_inorder();
}
}
Output for preorder would be:
1:5, 4, 3, 2,
3:7, 6,
5:8,
8:11, 10, 9,
Output for inorder would be:
3:7, 6,
8:11, 10, 9,
5:8,
1:5, 4, 3, 2,
Edit2: Just for completeness, the post-order traversal as well :-)
void traverse_postorder(void) {
if (this->left != NULL) {
this->left->traverse_postorder();
}
if (this->right != NULL) {
this->right->traverse_postorder();
}
if (this->left != NULL) {
cout << data << ":";
this->left->print_siblings();
cout << endl;
}
}
And its output:
8:11, 10, 9,
5:8,
3:7, 6,
1:5, 4, 3, 2,

Answer 2

Try something like: (Not tested yet)
printTree(TNode root) {
queue <TNode> proc;
proc.push(root);
while (!proc.empty()) {
TNode front = proc.front();
proc.pop();
// add front's children to a stack
stack <Tnode> temp;
Tnode current = front->left;
while (current != NULL) {
temp.push(current);
current = current->right;
}
// add the children to the back of queue in reverse order using the stack.
// at the same time, print.
printf ("%lld:", front->data);
while (!temp.empty()) {
proc.push(temp.top());
printf(" %lld", temp.top()->data);
temp.pop();
}
printf("\n");
}
}
I am still trying to make it more elegant. Thanks for the interesting problem!
Edit: Changed the format specifiers to lld

Answer 3

I've made something like this, but kind'a unrecursive due to while loops.
Is there any way making it more rec
void mirrorChilds(TNode * root)//
{
if(!root) return;
cout << root->data << " ";
//tmp = tmp->right;
if(!root->left) return;
TNode * tmp = root->left;
while(tmp != NULL)
{
root->left = tmp;
tmp = tmp->right;
}
tmp = root->left;
while(root != tmp)
{
cout << tmp->data << " ";
tmp = tmp->parent;
}
cout << endl;
tmp = root->left;
while(root != tmp)
{
if(tmp->left) mirrorChilds(tmp);
//if(tmp->left) cout << tmp->left->data << endl;
tmp = tmp->parent;
}
}
It works prefectly fine, but ye, i kinda trying to get O(n*log(n)) time ...