Deletion in a red-black tree is a bit more complicated than insertion. When a node is to be deleted, it can either have no children, one child or two children.
Here are the steps involved in deleting a node in a red-black tree:
- If the node to be deleted has no children, simply remove it and update the parent node.
- If the node to be deleted has only one child, replace the node with its child.
- If the node to be deleted has two children, then replace the node with its in-order successor, which is the leftmost node in the right subtree. Then delete the in-order successor node as if it has at most one child.
- After the node is deleted, the red-black properties might be violated. To restore these properties, some color changes and rotations are performed on the nodes in the tree. The changes are similar to those performed during insertion, but with different conditions.
- The deletion operation in a red-black tree takes O(log n) time on average, making it a good choice for searching and deleting elements in large data sets.
Reference books for more information on Red-Black Trees and their implementation in various programming languages:
- “Introduction to Algorithms” by Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein
- “The Art of Computer Programming, Volume 3: Sorting and Searching” by Donald E. Knuth
- “Algorithms in C, Part 5: Graph Algorithms” by Robert Sedgewick.
We have discussed the following topics on the Red-Black tree in previous posts. We strongly recommend referring following post as a prerequisite of this post.
Red-Black Tree Introduction
Red Black Tree Insert
Insertion Vs Deletion:
Like Insertion, recoloring and rotations are used to maintain the Red-Black properties.
In the insert operation, we check the color of the uncle to decide the appropriate case. In the delete operation, we check the color of the sibling to decide the appropriate case.
The main property that violates after insertion is two consecutive reds. In delete, the main violated property is, change of black height in subtrees as deletion of a black node may cause reduced black height in one root to leaf path.
Deletion is a fairly complex process. To understand deletion, the notion of double black is used. When a black node is deleted and replaced by a black child, the child is marked as double black. The main task now becomes to convert this double black to single black.
Deletion Steps
Following are detailed steps for deletion.
1) Perform standard BST delete. When we perform standard delete operation in BST, we always end up deleting a node which is an either leaf or has only one child (For an internal node, we copy the successor and then recursively call delete for successor, successor is always a leaf node or a node with one child). So we only need to handle cases where a node is leaf or has one child. Let v be the node to be deleted and u be the child that replaces v (Note that u is NULL when v is a leaf and color of NULL is considered as Black).
2) Simple Case: If either u or v is red, we mark the replaced child as black (No change in black height). Note that both u and v cannot be red as v is parent of u and two consecutive reds are not allowed in red-black tree.
3) If Both u and v are Black.
3.1) Color u as double black. Now our task reduces to convert this double black to single black. Note that If v is leaf, then u is NULL and color of NULL is considered black. So the deletion of a black leaf also causes a double black.
3.2) Do following while the current node u is double black, and it is not the root. Let sibling of node be s.
….(a): If sibling s is black and at least one of sibling’s children is red, perform rotation(s). Let the red child of s be r. This case can be divided in four subcases depending upon positions of s and r.
…………..(i) Left Left Case (s is left child of its parent and r is left child of s or both children of s are red). This is mirror of right right case shown in below diagram.
…………..(ii) Left Right Case (s is left child of its parent and r is right child). This is mirror of right left case shown in below diagram.
…………..(iii) Right Right Case (s is right child of its parent and r is right child of s or both children of s are red)
…………..(iv) Right Left Case (s is right child of its parent and r is left child of s)
…..(b): If sibling is black and its both children are black, perform recoloring, and recur for the parent if parent is black.
In this case, if parent was red, then we didn’t need to recur for parent, we can simply make it black (red + double black = single black)
…..(c): If sibling is red, perform a rotation to move old sibling up, recolor the old sibling and parent. The new sibling is always black (See the below diagram). This mainly converts the tree to black sibling case (by rotation) and leads to case (a) or (b). This case can be divided in two subcases.
…………..(i) Left Case (s is left child of its parent). This is mirror of right right case shown in below diagram. We right rotate the parent p.
…………..(ii) Right Case (s is right child of its parent). We left rotate the parent p.
3.3) If u is root, make it single black and return (Black height of complete tree reduces by 1).
below is the C++ implementation of above approach:
CPP
#include <iostream>#include <queue>using namespace std;enum COLOR { RED, BLACK };class Node {public: int val; COLOR color; Node *left, *right, *parent; Node(int val) : val(val) { parent = left = right = NULL; // Node is created during insertion // Node is red at insertion color = RED; } // returns pointer to uncle Node *uncle() { // If no parent or grandparent, then no uncle if (parent == NULL or parent->parent == NULL) return NULL; if (parent->isOnLeft()) // uncle on right return parent->parent->right; else // uncle on left return parent->parent->left; } // check if node is left child of parent bool isOnLeft() { return this == parent->left; } // returns pointer to sibling Node *sibling() { // sibling null if no parent if (parent == NULL) return NULL; if (isOnLeft()) return parent->right; return parent->left; } // moves node down and moves given node in its place void moveDown(Node *nParent) { if (parent != NULL) { if (isOnLeft()) { parent->left = nParent; } else { parent->right = nParent; } } nParent->parent = parent; parent = nParent; } bool hasRedChild() { return (left != NULL and left->color == RED) or (right != NULL and right->color == RED); }};class RBTree { Node *root; // left rotates the given node void leftRotate(Node *x) { // new parent will be node's right child Node *nParent = x->right; // update root if current node is root if (x == root) root = nParent; x->moveDown(nParent); // connect x with new parent's left element x->right = nParent->left; // connect new parent's left element with node // if it is not null if (nParent->left != NULL) nParent->left->parent = x; // connect new parent with x nParent->left = x; } void rightRotate(Node *x) { // new parent will be node's left child Node *nParent = x->left; // update root if current node is root if (x == root) root = nParent; x->moveDown(nParent); // connect x with new parent's right element x->left = nParent->right; // connect new parent's right element with node // if it is not null if (nParent->right != NULL) nParent->right->parent = x; // connect new parent with x nParent->right = x; } void swapColors(Node *x1, Node *x2) { COLOR temp; temp = x1->color; x1->color = x2->color; x2->color = temp; } void swapValues(Node *u, Node *v) { int temp; temp = u->val; u->val = v->val; v->val = temp; } // fix red red at given node void fixRedRed(Node *x) { // if x is root color it black and return if (x == root) { x->color = BLACK; return; } // initialize parent, grandparent, uncle Node *parent = x->parent, *grandparent = parent->parent, *uncle = x->uncle(); if (parent->color != BLACK) { if (uncle != NULL && uncle->color == RED) { // uncle red, perform recoloring and recurse parent->color = BLACK; uncle->color = BLACK; grandparent->color = RED; fixRedRed(grandparent); } else { // Else perform LR, LL, RL, RR if (parent->isOnLeft()) { if (x->isOnLeft()) { // for left right swapColors(parent, grandparent); } else { leftRotate(parent); swapColors(x, grandparent); } // for left left and left right rightRotate(grandparent); } else { if (x->isOnLeft()) { // for right left rightRotate(parent); swapColors(x, grandparent); } else { swapColors(parent, grandparent); } // for right right and right left leftRotate(grandparent); } } } } // find node that do not have a left child // in the subtree of the given node Node *successor(Node *x) { Node *temp = x; while (temp->left != NULL) temp = temp->left; return temp; } // find node that replaces a deleted node in BST Node *BSTreplace(Node *x) { // when node have 2 children if (x->left != NULL and x->right != NULL) return successor(x->right); // when leaf if (x->left == NULL and x->right == NULL) return NULL; // when single child if (x->left != NULL) return x->left; else return x->right; } // deletes the given node void deleteNode(Node *v) { Node *u = BSTreplace(v); // True when u and v are both black bool uvBlack = ((u == NULL or u->color == BLACK) and (v->color == BLACK)); Node *parent = v->parent; if (u == NULL) { // u is NULL therefore v is leaf if (v == root) { // v is root, making root null root = NULL; } else { if (uvBlack) { // u and v both black // v is leaf, fix double black at v fixDoubleBlack(v); } else { // u or v is red if (v->sibling() != NULL) // sibling is not null, make it red" v->sibling()->color = RED; } // delete v from the tree if (v->isOnLeft()) { parent->left = NULL; } else { parent->right = NULL; } } delete v; return; } if (v->left == NULL or v->right == NULL) { // v has 1 child if (v == root) { // v is root, assign the value of u to v, and delete u v->val = u->val; v->left = v->right = NULL; delete u; } else { // Detach v from tree and move u up if (v->isOnLeft()) { parent->left = u; } else { parent->right = u; } delete v; u->parent = parent; if (uvBlack) { // u and v both black, fix double black at u fixDoubleBlack(u); } else { // u or v red, color u black u->color = BLACK; } } return; } // v has 2 children, swap values with successor and recurse swapValues(u, v); deleteNode(u); } void fixDoubleBlack(Node *x) { if (x == root) // Reached root return; Node *sibling = x->sibling(), *parent = x->parent; if (sibling == NULL) { // No sibling, double black pushed up fixDoubleBlack(parent); } else { if (sibling->color == RED) { // Sibling red parent->color = RED; sibling->color = BLACK; if (sibling->isOnLeft()) { // left case rightRotate(parent); } else { // right case leftRotate(parent); } fixDoubleBlack(x); } else { // Sibling black if (sibling->hasRedChild()) { // at least 1 red children if (sibling->left != NULL and sibling->left->color == RED) { if (sibling->isOnLeft()) { // left left sibling->left->color = sibling->color; sibling->color = parent->color; rightRotate(parent); } else { // right left sibling->left->color = parent->color; rightRotate(sibling); leftRotate(parent); } } else { if (sibling->isOnLeft()) { // left right sibling->right->color = parent->color; leftRotate(sibling); rightRotate(parent); } else { // right right sibling->right->color = sibling->color; sibling->color = parent->color; leftRotate(parent); } } parent->color = BLACK; } else { // 2 black children sibling->color = RED; if (parent->color == BLACK) fixDoubleBlack(parent); else parent->color = BLACK; } } } } // prints level order for given node void levelOrder(Node *x) { if (x == NULL) // return if node is null return; // queue for level order queue<Node *> q; Node *curr; // push x q.push(x); while (!q.empty()) { // while q is not empty // dequeue curr = q.front(); q.pop(); // print node value cout << curr->val << " "; // push children to queue if (curr->left != NULL) q.push(curr->left); if (curr->right != NULL) q.push(curr->right); } } // prints inorder recursively void inorder(Node *x) { if (x == NULL) return; inorder(x->left); cout << x->val << " "; inorder(x->right); }public: // constructor // initialize root RBTree() { root = NULL; } Node *getRoot() { return root; } // searches for given value // if found returns the node (used for delete) // else returns the last node while traversing (used in insert) Node *search(int n) { Node *temp = root; while (temp != NULL) { if (n < temp->val) { if (temp->left == NULL) break; else temp = temp->left; } else if (n == temp->val) { break; } else { if (temp->right == NULL) break; else temp = temp->right; } } return temp; } // inserts the given value to tree void insert(int n) { Node *newNode = new Node(n); if (root == NULL) { // when root is null // simply insert value at root newNode->color = BLACK; root = newNode; } else { Node *temp = search(n); if (temp->val == n) { // return if value already exists return; } // if value is not found, search returns the node // where the value is to be inserted // connect new node to correct node newNode->parent = temp; if (n < temp->val) temp->left = newNode; else temp->right = newNode; // fix red red violation if exists fixRedRed(newNode); } } // utility function that deletes the node with given value void deleteByVal(int n) { if (root == NULL) // Tree is empty return; Node *v = search(n), *u; if (v->val != n) { cout << "No node found to delete with value:" << n << endl; return; } deleteNode(v); } // prints inorder of the tree void printInOrder() { cout << "Inorder: " << endl; if (root == NULL) cout << "Tree is empty" << endl; else inorder(root); cout << endl; } // prints level order of the tree void printLevelOrder() { cout << "Level order: " << endl; if (root == NULL) cout << "Tree is empty" << endl; else levelOrder(root); cout << endl; }};int main() { RBTree tree; tree.insert(7); tree.insert(3); tree.insert(18); tree.insert(10); tree.insert(22); tree.insert(8); tree.insert(11); tree.insert(26); tree.insert(2); tree.insert(6); tree.insert(13); tree.printInOrder(); tree.printLevelOrder(); cout<<endl<<"Deleting 18, 11, 3, 10, 22"<<endl; tree.deleteByVal(18); tree.deleteByVal(11); tree.deleteByVal(3); tree.deleteByVal(10); tree.deleteByVal(22); tree.printInOrder(); tree.printLevelOrder(); return 0;} |
Java
// Java Code for the above approachimport java.util.LinkedList;import java.util.Queue;enum COLOR { RED, BLACK }class Node { int val; COLOR color; Node left, right, parent; Node(int val) { this.val = val; parent = left = right = null; // Node is created during insertion // Node is red at insertion color = COLOR.RED; } // returns pointer to uncle Node uncle() { // If no parent or grandparent, then no uncle if (parent == null || parent.parent == null) return null; if (parent.isOnLeft()) // uncle on right return parent.parent.right; else // uncle on left return parent.parent.left; } // check if node is left child of parent boolean isOnLeft() { return this == parent.left; } // returns pointer to sibling Node sibling() { // sibling null if no parent if (parent == null) return null; if (isOnLeft()) return parent.right; return parent.left; } // moves node down and moves given node in its place void moveDown(Node nParent) { if (parent != null) { if (isOnLeft()) parent.left = nParent; else parent.right = nParent; } nParent.parent = parent; parent = nParent; } boolean hasRedChild() { return (left != null && left.color == COLOR.RED) || (right != null && right.color == COLOR.RED); }}class RBTree { Node root; // left rotates the given node void leftRotate(Node x) { // new parent will be node's right child Node nParent = x.right; // update root if current node is root if (x == root) root = nParent; x.moveDown(nParent); // connect x with new parent's left element x.right = nParent.left; // connect new parent's left element with node // if it is not null if (nParent.left != null) nParent.left.parent = x; // connect new parent with x nParent.left = x; } void rightRotate(Node x) { // new parent will be node's left child Node nParent = x.left; // update root if current node is root if (x == root) root = nParent; x.moveDown(nParent); // connect x with new parent's right element x.left = nParent.right; // connect new parent's right element with node // if it is not null if (nParent.right != null) nParent.right.parent = x; // connect new parent with x nParent.right = x; } void swapColors(Node x1, Node x2) { COLOR temp = x1.color; x1.color = x2.color; x2.color = temp; } void swapValues(Node u, Node v) { int temp = u.val; u.val = v.val; v.val = temp; } // fix red red at given node void fixRedRed(Node x) { // if x is root color it black and return if (x == root) { x.color = COLOR.BLACK; return; } // initialize parent, grandparent, uncle Node parent = x.parent, grandparent = parent.parent, uncle = x.uncle(); if (parent.color != COLOR.BLACK) { if (uncle != null && uncle.color == COLOR.RED) { // uncle red, perform recoloring and recurse parent.color = COLOR.BLACK; uncle.color = COLOR.BLACK; grandparent.color = COLOR.RED; fixRedRed(grandparent); } else { // Else perform LR, LL, RL, RR if (parent.isOnLeft()) { if (x.isOnLeft()) // for left right swapColors(parent, grandparent); else { leftRotate(parent); swapColors(x, grandparent); } // for left left and left right rightRotate(grandparent); } else { if (x.isOnLeft()) { // for right left rightRotate(parent); swapColors(x, grandparent); } else swapColors(parent, grandparent); // for right right and right left leftRotate(grandparent); } } } } // find node that do not have a left child // in the subtree of the given node Node successor(Node x) { Node temp = x; while (temp.left != null) temp = temp.left; return temp; } // find node that replaces a deleted node in BST Node BSTreplace(Node x) { // when node have 2 children if (x.left != null && x.right != null) return successor(x.right); // when leaf if (x.left == null && x.right == null) return null; // when single child if (x.left != null) return x.left; else return x.right; } // deletes the given node void deleteNode(Node v) { Node u = BSTreplace(v); // True when u and v are both black boolean uvBlack = ((u == null || u.color == COLOR.BLACK) && (v.color == COLOR.BLACK)); Node parent = v.parent; if (u == null) { // u is NULL therefore v is leaf if (v == root) // v is root, making root null root = null; else { if (uvBlack) // u and v both black // v is leaf, fix double black at v fixDoubleBlack(v); // u or v is red else if (v.sibling() != null) // sibling is not null, make it red v.sibling().color = COLOR.RED; // delete v from the tree if (v.isOnLeft()) parent.left = null; else parent.right = null; } return; } if (v.left == null || v.right == null) { // v has 1 child if (v == root) { // v is root, assign the value of u to v, and delete u v.val = u.val; v.left = v.right = null; // delete u; } else { // Detach v from tree and move u up if (v.isOnLeft()) parent.left = u; else parent.right = u; u.parent = parent; if (uvBlack) // u and v both black, fix double black at u fixDoubleBlack(u); else // u or v red, color u black u.color = COLOR.BLACK; } return; } // v has 2 children, swap values with successor and recurse swapValues(u, v); deleteNode(u); } void fixDoubleBlack(Node x) { // Reached root if (x == root) return; Node sibling = x.sibling(), parent = x.parent; if (sibling == null) // No sibling, double black pushed up fixDoubleBlack(parent); else { if (sibling.color == COLOR.RED) { // sibling red parent.color = COLOR.RED; sibling.color = COLOR.BLACK; if (sibling.isOnLeft()) // right case rightRotate(parent); else // right case leftRotate(parent); fixDoubleBlack(x); } else { // Sibling black if (sibling.hasRedChild()) { // at least 1 red children if (sibling.left != null && sibling.left.color == COLOR.RED) { if (sibling.isOnLeft()) { // left left sibling.left.color = sibling.color; sibling.color = parent.color; rightRotate(parent); } else { // right right sibling.left.color = parent.color; rightRotate(sibling); leftRotate(parent); } } else { if (sibling.isOnLeft()) { // left right sibling.right.color = parent.color; leftRotate(sibling); rightRotate(parent); } else { // right right sibling.right.color = sibling.color; sibling.color = parent.color; leftRotate(parent); } } parent.color = COLOR.BLACK; } else { // 2 black children sibling.color = COLOR.RED; if (parent.color == COLOR.BLACK) fixDoubleBlack(parent); else parent.color = COLOR.BLACK; } } } } // prints level order for given node void levelOrder(Node x) { if (x == null) return; // queue for level order Queue<Node> q = new LinkedList<>(); Node curr; q.add(x); while (!q.isEmpty()) { curr = q.poll(); // print node value System.out.print(curr.val + " "); // push children to queue if (curr.left != null) q.add(curr.left); if (curr.right != null) q.add(curr.right); } } // prints inorder recursively void inorder(Node x) { if (x == null) return; inorder(x.left); System.out.print(x.val + " "); inorder(x.right); } // constructor // initialize root RBTree() { root = null; } Node getRoot() { return root; } // searches for given value // if found returns the node (used for delete) // else returns the last node while traversing (used in insert) Node search(int n) { Node temp = root; while (temp != null) { if (n < temp.val) { if (temp.left == null) break; else temp = temp.left; } else if (n == temp.val) { break; } else { if (temp.right == null) break; else temp = temp.right; } } return temp; } // inserts the given value to tree void insert(int n) { Node newNode = new Node(n); if (root == null) { // when root is null // simply insert value at root newNode.color = COLOR.BLACK; root = newNode; } else { Node temp = search(n); // return if value already exists if (temp.val == n) return; // if value is not found, search returns the node // where the value is to be inserted // connect new node to correct node newNode.parent = temp; if (n < temp.val) temp.left = newNode; else temp.right = newNode; // fix red red violation if exists fixRedRed(newNode); } } // utility function that deletes the node with given value void deleteByVal(int n) { if (root == null) return; Node v = search(n), u; if (v.val != n) { System.out.println("No node found to delete with value: " + n); return; } deleteNode(v); } // prints inorder of the tree void printInOrder() { System.out.println("Inorder: "); if (root == null) System.out.println("Tree is empty"); else inorder(root); System.out.println(); } // prints level order of the tree void printLevelOrder() { System.out.println("Level order: "); if (root == null) System.out.println("Tree is empty"); else levelOrder(root); System.out.println(); }}public class Main { public static void main(String[] args) { RBTree tree = new RBTree(); tree.insert(7); tree.insert(3); tree.insert(18); tree.insert(10); tree.insert(22); tree.insert(8); tree.insert(11); tree.insert(26); tree.insert(2); tree.insert(6); tree.insert(13); tree.printInOrder(); tree.printLevelOrder(); System.out.println("\nDeleting 18, 11, 3, 10, 22\n"); tree.deleteByVal(18); tree.deleteByVal(11); tree.deleteByVal(3); tree.deleteByVal(10); tree.deleteByVal(22); tree.printInOrder(); tree.printLevelOrder(); }}// This code is contributed by Abhinav Mahajan (abhinav_m22). |
Output:
Inorder:
2 3 6 7 8 10 11 13 18 22 26
Level order:
10 7 18 3 8 11 22 2 6 13 26
Deleting 18, 11, 3, 10, 22
Inorder:
2 6 7 8 13 26
Level order:
13 7 26 6 8 2
References:
https://www.cs.purdue.edu/homes/ayg/CS251/slides/chap13c.pdf
Introduction to Algorithms 3rd Edition by Clifford Stein, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest
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