Monday, 17 July 2017 03:55

Write a program to Delete a Tree.

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To delete a tree we must traverse all the nodes of the tree and delete them one by one. So which traversal we should use – Inorder or Preorder or Postorder. Answer is simple – Postorder, because before deleting the parent node we should delete its children nodes first

We can delete tree with other traversals also with extra space complexity but why should we go for other traversals if we have Postorder available which does the work without storing anything in same time complexity.

For the following tree nodes are deleted in order – 4, 5, 2, 3, 1

#include<stdio.h>
#include<stdlib.h>

/* 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);
}

/*  This function traverses tree in post order to
to delete each and every node of the tree */
void deleteTree(struct node* node)
{
if (node == NULL) return;

/* first delete both subtrees */
deleteTree(node->left);
deleteTree(node->right);

/* then delete the node */
printf("\n Deleting node: %d", node->data);
free(node);
}

/* Driver program to test deleteTree function*/
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);

deleteTree(root);
root = NULL;

printf("\n Tree deleted ");

getchar();
return 0;
}
Output:

Deleting node: 4
Deleting node: 5
Deleting node: 2
Deleting node: 3
Deleting node: 1
Tree deleted

The above deleteTree() function deletes the tree, but doesn’t change root to NULL which may cause problems if the user of deleteTree() doesn’t change root to NULL and tires to access values using root pointer. We can modify the deleteTree() function to take reference to the root node so that this problem doesn’t occur. See the following code.

#include<stdio.h>
#include<stdlib.h>

/* 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);
}

/*  This function is same as deleteTree() in the previous program */
void _deleteTree(struct node* node)
{
if (node == NULL) return;

/* first delete both subtrees */
_deleteTree(node->left);
_deleteTree(node->right);

/* then delete the node */
printf("\n Deleting node: %d", node->data);
free(node);
}

/* Deletes a tree and sets the root as NULL */
void deleteTree(struct node** node_ref)
{
_deleteTree(*node_ref);
*node_ref = NULL;
}

/* Driver program to test deleteTree function*/
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);

// Note that we pass the address of root here
deleteTree(&root);
printf("\n Tree deleted ");

getchar();
return 0;
}
Deleting node: 4
Deleting node: 5
Deleting node: 2
Deleting node: 3
Deleting node: 1
Tree deleted

Time Complexity: O(n)
Space Complexity: If we don’t consider size of stack for function calls then O(1) otherwise O(n) 