Deletion in Red-Black Tree

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:

1. If the node to be deleted has no children, simply remove it and update the parent node.
2. If the node to be deleted has only one child, replace the node with its child.
3. 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.
4. 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.
5. 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:

1. "Introduction to Algorithms" by Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein
2. "The Art of Computer Programming, Volume 3: Sorting and Searching" by Donald E. Knuth
3. "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. 

👁 rbdelete11

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.

👁 rbdelete12_new

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) 

👁 rbdelete13New

..............(iv) Right Left Case (s is right child of its parent and r is left child of s) 

👁 rbdelete14

.....(b): If sibling is black and its both children are black, perform recoloring, and recur for the parent if parent is black. 

👁 rbdelete15

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. 

👁 rbdelete16

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: 

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