WAP Convert Given Binary Tree into its Mirror Tree Efficiently

This Problem Can Be Solved Using Two Approach

1.Top Down
2 Bottom Up


1.Top Down
Start from The Root create copy of it & then create copy for left & right sub tree by calling copy function recursively


#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;
};

/* 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);
}

/* Change a tree so that the roles of the left and
right pointers are swapped at every node.

So the tree...
4
/ \
2 5
/ \
1 3

is changed to...
4
/ \
5 2
/ \
3 1
*/
struct node* copy(struct node *root)
{
struct node *temp;

if(root==NULL)
return(NULL);
temp = (struct node*) malloc(sizeof(struct node));
temp->data= root->data;

temp->left = copy(root->right);
temp->right = copy(root->left);

return(temp);
}

/* Helper function to test mirror(). Given a binary
search tree, print out its data elements in
increasing sorted order.*/
void inOrder(struct node* node)
{
if (node == NULL)
return;

inOrder(node->left);
printf("%d ", node->data);

inOrder(node->right);
}

/* Driver program to test mirror() */
int main()
{
struct node *root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
root->left->left->left = newNode(6);
root->left->right->right = newNode(7);
/* Print inorder traversal of the input tree */
printf("\n Inorder traversal of the constructed tree is \n");
inOrder(root);

/* Convert tree to its mirror */
root=copy(root);

/* Print inorder traversal of the mirror tree */
printf("\n Inorder traversal of the mirror tree is \n");
inOrder(root);

getchar();
return 0;
}

Time Complexity O(n)
Space Complexity O(1) if stack space not considered else O(n)
Run Here https://ideone.com/QnMjQ


Bottom Up Approach ..First create mirror tree for left & right sub tree them comes to root


#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;
};

/* 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);
}

/* Change a tree so that the roles of the left and
right pointers are swapped at every node.

So the tree...
4
/ \
2 5
/ \
1 3

is changed to...
4
/ \
5 2
/ \
3 1
*/
void copy(struct node* node)
{
if (node==NULL)
return;
else
{
struct node* temp;

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

/* swap the pointers in this node */
temp = node->left;
node->left = node->right;
node->right = temp;
}
}

/* Helper function to test mirror(). Given a binary
search tree, print out its data elements in
increasing sorted order.*/
void inOrder(struct node* node)
{
if (node == NULL)
return;

inOrder(node->left);
printf("%d ", node->data);

inOrder(node->right);
}

/* Driver program to test mirror() */
int main()
{
struct node *root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);

/* Print inorder traversal of the input tree */
printf("\n Inorder traversal of the constructed tree is \n");
inOrder(root);

/* Convert tree to its mirror */
copy(root);

/* Print inorder traversal of the mirror tree */
printf("\n Inorder traversal of the mirror tree is \n");
inOrder(root);

getchar();
return 0;
}


Time Complexity O(n)
Space Complexity O(1) if stack space not considered else O(n)
Run Here https://ideone.com/emkO7