Write a program that converts a given tree to its Double tree. To create Double tree of the given tree, create a new duplicate for each node, and insert the duplicate as the left child of the original node.

e.g below tree

1
/ \
2 3
/ \
4 5
is changed to

1
/ \
1 3
/ /
2 3
/ \
2 5
/ /
4 5
/
4

Algorithm:
Recursively convert the tree to double tree in postorder fashion. For each node, first convert the left subtree of the node, then right subtree, finally create a duplicate node of the node and fix the left child of the node and left child of left child.

#include
#include

/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct node
{
int data;
struct node* left;
struct node* right;
};

/* function to create a new node of tree and returns pointer */
struct node* newNode(int data);

/* Function to convert a tree to double tree */
void doubleTree(struct node* node)
{
struct node* oldLeft;

if (node==NULL) return;

/* do the subtrees */
doubleTree(node->left);
doubleTree(node->right);

/* duplicate this node to its left */
oldLeft = node->left;
node->left = newNode(node->data);
node->left->left = oldLeft;
}

/* UTILITY FUNCTIONS TO TEST doubleTree() FUNCTION */
/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
struct node* newNode(int data)
{
struct node* node = (struct node*)
malloc(sizeof(struct node));
node->data = data;
node->left = NULL;
node->right = NULL;

return(node);
}

/* Given a binary tree, print its nodes in inorder*/
void printInorder(struct node* node)
{
if (node == NULL)
return;
printInorder(node->left);
printf("%d ", node->data);
printInorder(node->right);
}

/* Driver program to test above functions*/
int main()
{

/* Constructed binary tree is
1
/ \
2 3
/ \
4 5
*/
struct node *root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);

printf("Inorder traversal of the original tree is \n");
printInorder(root);

doubleTree(root);

printf("\n Inorder traversal of the double tree is \n");
printInorder(root);

getchar();
return 0;
}
Time Complexity: O(n) where n is the number of nodes in the tree.