Red-Black Tree Deletion

http://www.geeksforgeeks.org/red-black-tree-set-3-delete-2/

Insertion Vs Deletion:

Like Insertion, recoloring and rotations are used to maintain the Red-Black properties.

In insert operation, we check color of uncle to decide the appropriate case. In delete operation,we check color of 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 fairly complex process. To understand deletion, 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 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 deleted and u be the child that replaces v.

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 as 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 or it is not 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.