Saturday, November 16, 2024
Google search engine
HomeData Modelling & AIC Program for Red Black Tree Insertion

C Program for Red Black Tree Insertion

Following article is extension of article discussed here.
In AVL tree insertion, we used rotation as a tool to do balancing after insertion caused imbalance. In Red-Black tree, we use two tools to do balancing.

  1. Recoloring 
  2. Rotation

We try recoloring first, if recoloring doesn’t work, then we go for rotation. Following is detailed algorithm. The algorithms has mainly two cases depending upon the color of uncle. If uncle is red, we do recoloring. If uncle is black, we do rotations and/or recoloring.
Color of a NULL node is considered as BLACK.

Let x be the newly inserted node. 

  1. Perform standard BST insertion and make the color of newly inserted nodes as RED.
  2. If x is root, change color of x as BLACK (Black height of complete tree increases by 1).
  3. Do following if color of x’s parent is not BLACK or x is not root. 
    1. If x’s uncle is RED (Grand parent must have been black from property 4
      1. Change color of parent and uncle as BLACK. 
      2. color of grand parent as RED. 
      3.     3. Change x = x’s grandparent, repeat steps 2 and 3 for new x. 
         

redBlackCase2

      2. If x’s uncle is BLACK, then there can be four configurations for x, x’s parent (p) and x’s grandparent (g) (This is similar to AVL Tree

  1. Determine the configuration:
    1. Left Case (p is left child of g and x is left child of p).
    2. Left Right Case (p is left child of g and x is right child of p).
    3. Right Right Case (Mirror of case a).
    4. Right Left Case (Mirror of case c).
  2. Change x = x’s parent, repeat steps 2 and 3 for new x.

Following are operations to be performed in four subcases when uncle is BLACK.

All four cases when Uncle is BLACK

Left Left Case (See g, p and x) 

redBlackCase3a

Left Right Case (See g, p and x) 

redBlackCase3b

Right Right Case (See g, p and x) 

redBlackCase3c

Right Left Case (See g, p and x) 

redBlackCase3d

Examples of Insertion 

Examples

Implementation:

C




/** C implementation for
    Red-Black Tree Insertion
    This code is provided by
    costheta_z **/
#include <stdio.h>
#include <stdlib.h>
 
// Structure to represent each
// node in a red-black tree
struct node {
    int d; // data
    int c; // 1-red, 0-black
    struct node* p; // parent
    struct node* r; // right-child
    struct node* l; // left child
};
 
// global root for the entire tree
struct node* root = NULL;
 
// function to perform BST insertion of a node
struct node* bst(struct node* trav,
                      struct node* temp)
{
    // If the tree is empty,
    // return a new node
    if (trav == NULL)
        return temp;
 
    // Otherwise recur down the tree
    if (temp->d < trav->d)
    {
        trav->l = bst(trav->l, temp);
        trav->l->p = trav;
    }
    else if (temp->d > trav->d)
    {
        trav->r = bst(trav->r, temp);
        trav->r->p = trav;
    }
 
    // Return the (unchanged) node pointer
    return trav;
}
 
// Function performing right rotation
// of the passed node
void rightrotate(struct node* temp)
{
    struct node* left = temp->l;
    temp->l = left->r;
    if (temp->l)
        temp->l->p = temp;
    left->p = temp->p;
    if (!temp->p)
        root = left;
    else if (temp == temp->p->l)
        temp->p->l = left;
    else
        temp->p->r = left;
    left->r = temp;
    temp->p = left;
}
 
// Function performing left rotation
// of the passed node
void leftrotate(struct node* temp)
{
    struct node* right = temp->r;
    temp->r = right->l;
    if (temp->r)
        temp->r->p = temp;
    right->p = temp->p;
    if (!temp->p)
        root = right;
    else if (temp == temp->p->l)
        temp->p->l = right;
    else
        temp->p->r = right;
    right->l = temp;
    temp->p = right;
}
 
// This function fixes violations
// caused by BST insertion
void fixup(struct node* root, struct node* pt)
{
    struct node* parent_pt = NULL;
    struct node* grand_parent_pt = NULL;
 
    while ((pt != root) && (pt->c != 0)
           && (pt->p->c == 1))
    {
        parent_pt = pt->p;
        grand_parent_pt = pt->p->p;
 
        /*  Case : A
             Parent of pt is left child
             of Grand-parent of
           pt */
        if (parent_pt == grand_parent_pt->l)
        {
 
            struct node* uncle_pt = grand_parent_pt->r;
 
            /* Case : 1
                The uncle of pt is also red
                Only Recoloring required */
            if (uncle_pt != NULL && uncle_pt->c == 1)
            {
                grand_parent_pt->c = 1;
                parent_pt->c = 0;
                uncle_pt->c = 0;
                pt = grand_parent_pt;
            }
 
            else {
 
                /* Case : 2
                     pt is right child of its parent
                     Left-rotation required */
                if (pt == parent_pt->r) {
                    leftrotate(parent_pt);
                    pt = parent_pt;
                    parent_pt = pt->p;
                }
 
                /* Case : 3
                     pt is left child of its parent
                     Right-rotation required */
                rightrotate(grand_parent_pt);
                int t = parent_pt->c;
                parent_pt->c = grand_parent_pt->c;
                grand_parent_pt->c = t;
                pt = parent_pt;
            }
        }
 
        /* Case : B
             Parent of pt is right
             child of Grand-parent of
           pt */
        else {
            struct node* uncle_pt = grand_parent_pt->l;
 
            /*  Case : 1
                The uncle of pt is also red
                Only Recoloring required */
            if ((uncle_pt != NULL) && (uncle_pt->c == 1))
            {
                grand_parent_pt->c = 1;
                parent_pt->c = 0;
                uncle_pt->c = 0;
                pt = grand_parent_pt;
            }
            else {
                /* Case : 2
                   pt is left child of its parent
                   Right-rotation required */
                if (pt == parent_pt->l) {
                    rightrotate(parent_pt);
                    pt = parent_pt;
                    parent_pt = pt->p;
                }
 
                /* Case : 3
                     pt is right child of its parent
                     Left-rotation required */
                leftrotate(grand_parent_pt);
                int t = parent_pt->c;
                parent_pt->c = grand_parent_pt->c;
                grand_parent_pt->c = t;
                pt = parent_pt;
            }
        }
    }
}
 
// Function to print inorder traversal
// of the fixated tree
void inorder(struct node* trav)
{
    if (trav == NULL)
        return;
    inorder(trav->l);
    printf("%d ", trav->d);
    inorder(trav->r);
}
 
// driver code
int main()
{
    int n = 7;
    int a[7] = { 7, 6, 5, 4, 3, 2, 1 };
 
    for (int i = 0; i < n; i++) {
 
        // allocating memory to the node and initializing:
        // 1. color as red
        // 2. parent, left and right pointers as NULL
        // 3. data as i-th value in the array
        struct node* temp
            = (struct node*)malloc(sizeof(struct node));
        temp->r = NULL;
        temp->l = NULL;
        temp->p = NULL;
        temp->d = a[i];
        temp->c = 1;
 
        // calling function that performs bst insertion of
        // this newly created node
        root = bst(root, temp);
 
        // calling function to preserve properties of rb
        // tree
        fixup(root, temp);
          root->c = 0;
    }
 
    printf("Inorder Traversal of Created Tree\n");
    inorder(root);
 
    return 0;
}


Output

Inorder Traversal of Created Tree
1  2  3  4  5  6  7  

Level Order Traversal of Created Tree
6  4  7  2  5  1  3  

Time Complexity:  O(log n), as the height of red-black tree is O(log n) at most, and complexity of rotation is constant.
Auxiliary Space:  O(n),  here ‘n’ is the number of nodes in the red-black trees.

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!

RELATED ARTICLES

Most Popular

Recent Comments