Deletion in Splay Tree

It is recommended to refer following post as prerequisite of this post. Splay Tree | Set 1 (Search) Following are the different cases to delete a key k from splay tree.

1. If Root is NULL: We simply return the root.
2. Else Splay the given key k. If k is present, then it becomes the new root. If not present, then last accessed leaf node becomes the new root.
3. If new root’s key is not same as k, then return the root as k is not present.
4. Else the key k is present.
   • Split the tree into two trees Tree1 = root’s left subtree and Tree2 = root’s right subtree and delete the root node.
   • Let the root’s of Tree1 and Tree2 be Root1 and Root2 respectively.
   • If Root1 is NULL: Return Root2.
   • Else, Splay the maximum node (node having the maximum value) of Tree1.
   • After the Splay procedure, make Root2 as the right child of Root1 and return Root1.

Implementation:

CPP

// C implementation to delete a node from Splay Tree
#include <stdio.h>
#include <stdlib.h>

// An AVL tree node
struct node {
    int key;
    struct node *left, *right;
};

/* Helper function that allocates a new node with the given
   key and NULL left and right pointers. */
struct node* newNode(int key)
{
    struct node* node
        = (struct node*)malloc(sizeof(struct node));
    node->key = key;
    node->left = node->right = NULL;
    return (node);
}

// A utility function to right rotate subtree rooted with y
// See the diagram given above.
struct node* rightRotate(struct node* x)
{
    struct node* y = x->left;
    x->left = y->right;
    y->right = x;
    return y;
}

// A utility function to left rotate subtree rooted with x
// See the diagram given above.
struct node* leftRotate(struct node* x)
{
    struct node* y = x->right;
    x->right = y->left;
    y->left = x;
    return y;
}

// This function brings the key at root if key is present in
// tree. If key is not present, then it brings the last
// accessed item at root. This function modifies the tree
// and returns the new root
struct node* splay(struct node* root, int key)
{
    // Base cases: root is NULL or key is present at root
    if (root == NULL || root->key == key)
        return root;

    // Key lies in left subtree
    if (root->key > key) {
        // Key is not in tree, we are done
        if (root->left == NULL)
            return root;

        // Zig-Zig (Left Left)
        if (root->left->key > key) {
            // First recursively bring the key as root of
            // left-left
            root->left->left = splay(root->left->left, key);

            // Do first rotation for root, second rotation
            // is done after else
            root = rightRotate(root);
        }
        else if (root->left->key
                 < key) // Zig-Zag (Left Right)
        {
            // First recursively bring the key as root of
            // left-right
            root->left->right
                = splay(root->left->right, key);

            // Do first rotation for root->left
            if (root->left->right != NULL)
                root->left = leftRotate(root->left);
        }

        // Do second rotation for root
        return (root->left == NULL) ? root
                                    : rightRotate(root);
    }
    else // Key lies in right subtree
    {
        // Key is not in tree, we are done
        if (root->right == NULL)
            return root;

        // Zag-Zig (Right Left)
        if (root->right->key > key) {
            // Bring the key as root of right-left
            root->right->left
                = splay(root->right->left, key);

            // Do first rotation for root->right
            if (root->right->left != NULL)
                root->right = rightRotate(root->right);
        }
        else if (root->right->key
                 < key) // Zag-Zag (Right Right)
        {
            // Bring the key as root of right-right and do
            // first rotation
            root->right->right
                = splay(root->right->right, key);
            root = leftRotate(root);
        }

        // Do second rotation for root
        return (root->right == NULL) ? root
                                     : leftRotate(root);
    }
}

// The delete function for Splay tree. Note that this
// function returns the new root of Splay Tree after
// removing the key
struct node* delete_key(struct node* root, int key)
{
    struct node* temp;
    if (!root)
        return NULL;

    // Splay the given key
    root = splay(root, key);

    // If key is not present, then
    // return root
    if (key != root->key)
        return root;

    // If key is present
    // If left child of root does not exist
    // make root->right as root
    if (!root->left) {
        temp = root;
        root = root->right;
    }

    // Else if left child exits
    else {
        temp = root;

        /*Note: Since key == root->key,
        so after Splay(key, root->lchild),
        the tree we get will have no right child tree
        and maximum node in left subtree will get splayed*/
        // New root
        root = splay(root->left, key);

        // Make right child of previous root as
        // new root's right child
        root->right = temp->right;
    }

    // free the previous root node, that is,
    // the node containing the key
    free(temp);

    // return root of the new Splay Tree
    return root;
}

// A utility function to print preorder traversal of the
// tree. The function also prints height of every node
void preOrder(struct node* root)
{
    if (root != NULL) {
        printf("%d ", root->key);
        preOrder(root->left);
        preOrder(root->right);
    }
}

/* Driver program to test above function*/
int main()
{
    // Splay Tree Formation
    struct node* root = newNode(6);
    root->left = newNode(1);
    root->right = newNode(9);
    root->left->right = newNode(4);
    root->left->right->left = newNode(2);
    root->right->left = newNode(7);

    int key = 4;

    root = delete_key(root, key);
    printf("Preorder traversal of the modified Splay tree "
           "is \n");
    preOrder(root);
    return 0;
}

Output

Preorder traversal of the modified Splay tree is 
2 1 6 9 7 

References: https://www.geeksforgeeks.org/splay-tree-set-1-insert/ http://courses.cs.washington.edu/courses/cse326/01au/lectures/SplayTrees.ppt

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