CareerCup

It's not mirror image.

4 Answers
void Func(tNode **node){
if(*node!=NULL){
Func(&(*node)->pLeft);
tNode *temp;
temp = (*node)->pLeft;
(*node)->pLeft= (*node)->pRight;
(*node)->pRight = temp;
Func(&(*node)->pLeft);
}
}
will create a mirror image.
The above code will just swap root node's Left and Right child.

It creates a reflection of the tree as stated above

thank u guys bt
Let the input given be
98,15,100,10,78,120,5,12,96,110

What would be the output of the following code snippet?

int main(void)
{
tree *bT = new tree;
int i = 10;
int data;
while(i--){
std::cout<<"Enter the node"<<std::endl;
std::cin>>data;
bT->InsertNode(bT->GetRootNode(), data);


}

bT->Func(bT->GetRootNode());
bT->InsertNode(bT->GetRootNode(), 99);
bT->Func(bT->GetRootNode());
bT->traverse(bT->GetRootNode());
}
// insert() is simple bst insert and traverse() is postorder traversal

It absolutely does not creates an mirror image of the tree...

1. Imagine each parent node as a hinge about which child nodes rotate.
2. First, the leaf nodes rotate about their parent node. Then the level above leaf nodes rotate about their parents and so on.

It creates a mirror of tree , or swaps the children.