Red–black tree


In computer science, a red–black tree is a self-balancing binary search tree data structure noted for fast storage and retrieval of ordered information. The nodes in a red-black tree hold an extra "color" bit, often drawn as red and black, which help ensure that the tree is always approximately balanced.
When the tree is modified, the new tree is rearranged and "repainted" to restore the coloring properties that constrain how unbalanced the tree can become in the worst case. The properties are designed such that this rearranging and recoloring can be performed efficiently.
The balancing is not perfect, but guarantees searching in time, where is the number of entries in the tree. The [|insert] and delete operations, along with tree rearrangement and recoloring, also execute in time.
Tracking the color of each node requires only one bit of information per node because there are only two colors. The tree does not contain any other data specific to it being a red–black tree, so its memory footprint is almost identical to that of a classic binary search tree. In some cases, the added bit of information can be stored at no added memory cost.

History

In 1972, Rudolf Bayer invented a data structure that was a special order-4 case of a B-tree. These trees maintained all paths from root to leaf with the same number of nodes, creating perfectly balanced trees. However, they were not binary search trees. Bayer called them a "symmetric binary B-tree" in his paper and later they became popular as 2–3–4 trees or even 2–3 trees.
In a 1978 paper, "A Dichromatic Framework for Balanced Trees", Leonidas J. Guibas and Sedgewick (computer scientist)|Robert Sedgewick] derived the red–black tree from the symmetric binary B-tree. The color "red" was chosen because it was the best-looking color produced by the color laser printer available to the authors while working at Xerox PARC. Another response from Guibas states that it was because of the red and black pens available to them to draw the trees.
In 1993, Arne Andersson introduced the idea of a right leaning tree to simplify insert and delete operations.
In 1999, Chris Okasaki showed how to make the insert operation purely functional. Its balance function needed to take care of only four unbalanced cases and one default balanced case.
The original algorithm used eight unbalanced cases, but reduced that to six unbalanced cases. Sedgewick showed that the insert operation can be implemented in just 46 lines of Java.
In 2008, Sedgewick proposed the left-leaning red–black tree, leveraging Andersson’s idea that simplified the insert and delete operations. Sedgewick originally allowed nodes whose two children are red, making his trees more like 2–3–4 trees, but later this restriction was added, making new trees more like 2–3 trees. Sedgewick implemented the insert algorithm in just 33 lines, significantly shortening his original 46 lines of code.

Terminology

The black depth of a node is defined as the number of black nodes from the root to that node. The black height of a red–black tree is the number of black nodes in any path from the root to the leaves, which, by requirement 4, is constant.
The black height of a node is the black height of the subtree rooted by it. In this article, the black height of a null node shall be set to 0, because its subtree is empty as suggested by the example figure, and its tree height is also 0.

Properties

In addition to the requirements imposed on a binary search tree the following must be satisfied by a
  1. Every node is either red or black.
  2. All null nodes are considered black.
  3. A red node does not have a red child.
  4. Every path from a given node to any of its leaf nodes goes through the same number of black nodes.
  5. If a node N has exactly one child, the child must be red. If the child were black, its leaves would sit at a different black depth than N's null node, violating requirement 4.
Some authors, e.g. Cormen & al., claim "the root is black" as fifth requirement; but not Mehlhorn & Sanders or Sedgewick & Wayne. Since the root can always be changed from red to black, this rule has little effect on analysis.
This article also omits it, because it slightly disturbs the recursive algorithms and proofs.
As an example, every perfect binary tree that consists only of black nodes is a red–black tree.
The read-only operations, such as search or tree traversal, do not affect any of the requirements. In contrast, the modifying operations insert and delete easily maintain requirements 1 and 2, but with respect to the other requirements some extra effort must be made, to avoid introducing a violation of [|requirement 3], called a red-violation, or of [|requirement 4], called a black-violation.
The requirements enforce a critical property of red–black trees: the path from the root to the farthest leaf is no more than twice as long as the path from the root to the nearest leaf. The result is that the tree is height-balanced. Since operations such as inserting, deleting, and finding values require worst-case time proportional to the height of the tree, this upper bound on the height allows red–black trees to be efficient in the worst case, namely logarithmic in the number of entries, i.e. . For a mathematical proof see section Proof of bounds.
Red–black trees, like all binary search trees, allow quite efficient sequential access of their elements. But they support also asymptotically optimal direct access via a traversal from root to leaf, resulting in search time.

Analogy to 2–3–4 trees

Red–black trees are similar in structure to 2–3–4 trees, which are B-trees of order 4. In 2–3–4 trees, each node can contain between 1 and 3 values and have between 2 and 4 children. These 2–3–4 nodes correspond to black node – red children groups in red-black trees, as shown in figure 1. It is not a 1-to-1 correspondence, because 3-nodes have two equivalent representations: the red child may lie either to the left or right. The left-leaning red-black tree variant makes this relationship exactly 1-to-1, by only allowing the left child representation. Since every 2–3–4 node has a corresponding black node, invariant 4 of red-black trees is equivalent to saying that the leaves of a 2–3–4 tree all lie at the same level.
Despite structural similarities, operations on red–black trees are more economical than B-trees. B-trees require management of vectors of variable length, whereas red-black trees are simply binary trees.

Applications and related data structures

Red–black trees offer worst-case guarantees for insertion time, deletion time, and search time. Not only does this make them valuable in time-sensitive applications such as real-time applications, but it makes them valuable building blocks in other data structures that provide worst-case guarantees. For example, many data structures used in computational geometry are based on red–black trees, and the Completely Fair Scheduler and epoll system call of the Linux kernel use red–black trees.
The AVL tree is another structure supporting search, insertion, and removal. AVL trees can be colored red–black, and thus are a subset of red-black trees. The worst-case height of AVL is 0.720 times the worst-case height of red-black trees, so AVL trees are more rigidly balanced. The performance measurements of Ben Pfaff with realistic test cases in 79 runs find AVL to RB ratios between 0.677 and 1.077, median at 0.947, and geometric mean 0.910. The performance of WAVL trees lie in between AVL trees and red-black trees.
Red–black trees are also particularly valuable in functional programming, where they are one of the most common persistent data structures, used to construct associative arrays and sets that can retain previous versions after mutations. The persistent version of red–black trees requires space for each insertion or deletion, in addition to time.
For every 2–3–4 tree, there are corresponding red–black trees with data elements in the same order. The insertion and deletion operations on 2–3–4 trees are also equivalent to color-flipping and rotations in red–black trees. This makes 2–3–4 trees an important tool for understanding the logic behind red–black trees, and this is why many introductory algorithm texts introduce 2–3–4 trees just before red–black trees, even though 2–3–4 trees are not often used in practice.
In 2008, Sedgewick introduced a simpler version of the red–black tree called the left-leaning red–black tree by eliminating a previously unspecified degree of freedom in the implementation. The LLRB maintains an additional invariant that all red links must lean left except during inserts and deletes. Red–black trees can be made isometric to either 2–3 trees, or 2–3–4 trees, for any sequence of operations. The 2–3–4 tree isometry was described in 1978 by Sedgewick. With 2–3–4 trees, the isometry is resolved by a "color flip," corresponding to a split, in which the red color of two children nodes leaves the children and moves to the parent node.
The original description of the tango tree, a type of tree optimised for fast searches, specifically uses red–black trees as part of its data structure.
As of Java 8, the HashMap has been modified such that instead of using a LinkedList to store different elements with colliding hashcodes, a red–black tree is used. This results in the improvement of time complexity of searching such an element from to where is the number of elements with colliding hashes.

Implementation

The read-only operations, such as search or tree traversal, on a red–black tree require no modification from those used for binary search trees, because every red–black tree is a special case of a simple binary search tree. However, the immediate result of an insertion or removal may violate the properties of a red–black tree, the restoration of which is called rebalancing so that red–black trees become self-balancing.
Rebalancing has a worst-case time complexity of and average of, though these are very quick in practice. Additionally, rebalancing takes no more than three tree rotations.
This is an example implementation of insert and remove in C. Below are the data structures and the rotate_subtree helper function used in the insert and remove examples.

typedef enum Color: char Color;
typedef enum Direction: char Direction;
// red-black tree node
typedef struct Node Node;
typedef struct Tree;
static Direction direction
Node* rotate_subtree

Insertion

Insertion begins by placing the new node, say N, at the position in the binary search tree of a NULL node whose in-order predecessor’s key compares less than the new node’s key, which in turn compares less than the key of its in-order successor.


The rebalancing loop of the insert operation has the following invariants:Node is the current node, initially the insertion node.Node is red at the beginning of each iteration.
  • Requirement 3 is satisfied for all pairs node←parent with the possible exception node←'parent when parent' is also red.
  • All other properties are satisfied throughout the tree.

Insert case 1

The current node’s parent P is black, so requirement 3 holds. [|Requirement 4] holds also according to the loop invariant.

Insert case 2

If both the parent P and the uncle U are red, then both of them can be repainted black and the grandparent G becomes red for maintaining requirement 4. Since any path through the parent or uncle must pass through the grandparent, the number of black nodes on these paths has not changed. However, the grandparent G may now violate requirement 3, if it has a red parent. After relabeling G to N the loop invariant is fulfilled so that the rebalancing can be iterated on one black level higher.

Insert case 3

Insert case 2 has been executed for times and the total height of the tree has increased by 1, now being  .
The current node N is the root of the tree, and all RB-properties are satisfied.

Insert case 4

The parent P is red and the root.
Because N is also red, requirement 3 is violated. But after switching P’s color the tree is in RB-shape.
The black height of the tree increases by 1.

Insert case 5

The parent P is red but the uncle U is black. The ultimate goal is to rotate the parent node P to the grandparent position, but this will not work if N is an "inner" grandchild of G. A at P switches the roles of the current node N and its parent P. The rotation adds paths through N and removes paths through P. But both P and N are red, so requirement 4 is preserved. [|Requirement 3] is restored in case 6.

Insert case 6

The current node N is now certain to be an "outer" grandchild of G. Now at G, putting P in place of G and making P the parent of N and G. G is black and its former child P is red, since requirement 3 was violated. After switching the colors of P and G the resulting tree satisfies requirement 3. Requirement 4 also remains satisfied, since all paths that went through the black G now go through the black P.
Because the algorithm transforms the input without using an auxiliary data structure and using only a small amount of extra storage space for auxiliary variables it is in-place.

Removal

Simple cases

  • When the deleted node has 2 children, then we can swap its value with its in-order successor, and then delete the successor instead. Since the successor is leftmost, it can only have a right child or no child at all.
  • When the deleted node has only 1 child. In this case, just replace the node with its child, and color it black.
  • * The single child must be red according to conclusion 5, and the deleted node must be black according to requirement 3.
  • When the deleted node has no children and is the root, replace it with NULL. The tree is empty.
  • When the deleted node has no children, and is red, simply remove the leaf node.
  • When the deleted node has no children, and is black, deleting it will create an imbalance, and requires a rebalance, as covered in the next section.

Removal of a black non-root leaf

The complex case is when N is not the root, colored black and has no proper child.
In the first iteration, N is replaced by NULL.

void remove
The rebalancing loop of the delete operation has the following invariant:
  • At the beginning of each iteration the black height of N equals the iteration number minus one, which means that in the first iteration it is zero and that N is a true black node in higher iterations.
  • The number of black nodes on the paths through N is one less than before the deletion, whereas it is unchanged on all other paths, so that there is a black-violation at P if other paths exist.
  • All other properties are satisfied throughout the tree.

Delete case 1

The current node N is the new root. One black node has been removed from every path, so the RB-properties are preserved.
The black height of the tree decreases by 1.

Delete case 2

P, S, and S’s children are black. After painting S red all paths passing through S, which are precisely those paths not passing through N, have one less black node. Now all paths in the subtree rooted by P have the same number of black nodes, but one fewer than the paths that do not pass through P, so requirement 4 may still be violated. After relabeling P to N the loop invariant is fulfilled so that the rebalancing can be iterated on one black level higher.

Delete case 3

The sibling S is red, so P and the nephews C and D have to be black. A at P turns S into N’s grandparent.
Then after reversing the colors of P and S, the path through N is still short one black node. But N now has a red parent P and after the reassignment a black sibling S, so the transformations in cases 4, 5, or 6 are able to restore the RB-shape.

Delete case 4

The sibling S and S’s children are black, but P is red. Exchanging the colors of S and P does not affect the number of black nodes on paths going through S, but it does add one to the number of black nodes on paths going through N, making up for the deleted black node on those paths.

Delete case 5

The sibling S is black, S’s close child C is red, and S’s distant child D is black. After a at S the nephew C becomes S’s parent and N’s new sibling. The colors of S and C are exchanged.
All paths still have the same number of black nodes, but now N has a black sibling whose distant child is red, so the constellation is fit for case D6. Neither N nor its parent P are affected by this transformation, and P may be red or black.

Delete case 6

The sibling S is black, S’s distant child D is red. After a at P the sibling S becomes the parent of P and S’s distant child D. The colors of P and S are exchanged, and D is made black. The whole subtree still has the same color at its root S, namely either red or black, which refers to the same color both before and after the transformation. This way requirement 3 is preserved. The paths in the subtree not passing through N pass through the same number of black nodes as before, but N now has one additional black ancestor: either P has become black, or it was black and S was added as a black grandparent. Thus, the paths passing through N pass through one additional black node, so that requirement 4 is restored and the total tree is in RB-shape.
Because the algorithm transforms the input without using an auxiliary data structure and using only a small amount of extra storage space for auxiliary variables it is in-place.

Proof of bounds

For there is a red–black tree of height with
nodes and there is no red–black tree of this tree height with fewer nodes—therefore it is minimal.
Its black height is     or for odd also  
;Proof
For a red–black tree of a certain height to have minimal number of nodes, it must have exactly one longest path with maximal number of red nodes, to achieve a maximal tree height with a minimal black height. Besides this path all other nodes have to be black. If a node is taken off this tree it either loses height or some RB property.
The RB tree of height with red root is minimal. This is in agreement with
A minimal RB tree of height has a root whose two child subtrees are of different height. The higher child subtree is also a minimal RB tree, containing also a longest path that defines its height it has nodes and the black height The other subtree is a perfect binary tree of height having black nodes—and no red node. Then the number of nodes is by induction
The graph of the function is convex and piecewise linear with breakpoints at where The function has been tabulated as A027383 for
;Solving the function for
The inequality leads to, which for odd leads to
So in both, the even and the odd case, is in the interval
with being the number of nodes.
;Conclusion
A red–black tree with nodes has tree height

Set operations and bulk operations

In addition to the single-element insert, delete and lookup operations, several set operations have been defined on union, intersection and set difference. Then fast bulk operations on insertions or deletions can be implemented based on these set functions. These set operations rely on two helper operations, Split and Join. With the new operations, the implementation of red–black trees can be more efficient and highly-parallelizable. In order to achieve its time complexities this implementation requires that the root is allowed to be either red or black, and that every node stores its own black height.Join: The function Join is on two red–black trees and and a key, where, i.e. all keys in are less than, and all keys in are greater than. It returns a tree containing all elements in, also as.Split: To split a red–black tree into two smaller trees, those smaller than key, and those larger than key, first draw a path from the root by inserting into the red–black tree. After this insertion, all values less than will be found on the left of the path, and all values greater than will be found on the right. By applying Join, all the subtrees on the left side are merged bottom-up using keys on the path as intermediate nodes from bottom to top to form the left tree, and the right part is symmetric.
The join algorithm is as follows:
function joinRightRB:
if and :
return Node
T'=Node
if and :
T'.right.right.color=black;
return rotateLeft
return T' /* T */
function joinLeftRB:
/* symmetric to joinRightRB */
function join:
if TL.blackHeight>TR.blackHeight:
T'=joinRightRB
if and :
T'.color=black
return T'
if TR.blackHeight>TL.blackHeight:
/* symmetric */
if and :
return Node
return Node
The split algorithm is as follows:
function split:
if 'return
if return
if :
= split
return
= split
return
The union of two red–black trees and representing sets and, is a red–black tree that represents. The following recursive function computes this union:
function union:
if t1 = NULL return t2
if t2 = NULL return t1
=split
proc1=start:
TL=union
proc2=start:
TR=union
wait all proc1,proc2
return' join
Here, split is presumed to return two trees: one holding the keys less its input key, one holding the greater keys.
The algorithm for intersection or difference is similar, but requires the Join2 helper routine that is the same as Join but without the middle key. Based on the new functions for union, intersection or difference, either one key or multiple keys can be inserted to or deleted from the red–black tree. Since Split calls Join but does not deal with the balancing criteria of red–black trees directly, such an implementation is usually called the "join-based" implementation.
The complexity of each of union, intersection and difference is for two red–black trees of sizes and. This complexity is optimal in terms of the number of comparisons. More importantly, since the recursive calls to union, intersection or difference are independent of each other, they can be executed in parallel with a parallel depth. When, the join-based implementation has the same computational directed acyclic graph as single-element insertion and deletion if the root of the larger tree is used to split the smaller tree.

Parallel algorithms

Parallel algorithms for constructing red–black trees from sorted lists of items can run in constant time or time, depending on the computer model, if the number of processors available is asymptotically proportional to the number of items where. Fast search, insertion, and deletion parallel algorithms are also known.
The join-based algorithms for red–black trees are parallel for bulk operations, including union, intersection, construction, filter, map-reduce, and so on.

Parallel bulk operations

Basic operations like insertion, removal or update can be parallelised by defining operations that process bulks of multiple elements. It is also possible to process bulks with several basic operations, for example bulks may contain elements to insert and also elements to remove from the tree.
The algorithms for bulk operations aren't just applicable to the red–black tree, but can be adapted to other sorted sequence data structures also, like the 2–3 tree, 2–3–4 tree and (a,b)-tree. In the following different algorithms for bulk insert will be explained, but the same algorithms can also be applied to removal and update. Bulk insert is an operation that inserts each element of a sequence into a tree.

Join-based

This approach can be applied to every sorted sequence data structure that supports efficient join- and split-operations.
The general idea is to split and in multiple parts and perform the insertions on these parts in parallel.
  1. First the bulk of elements to insert must be sorted.
  2. After that, the algorithm splits into parts of about equal sizes.
  3. Next the tree must be split into parts in a way, so that the ranges of and overlap for corresponding parts only ; in other words, thanks to the ordering, for every following constraints hold:
  4. #
  5. #
  6. Now the algorithm inserts each element of into sequentially. This step must be performed for every, which can be done by up to processors in parallel.
  7. Finally, the resulting trees will be joined to form the final result of the entire operation.
Note that in Step 3 the constraints for splitting assure that in Step 5 the trees can be joined again and the resulting sequence is sorted.
The pseudo code shows a simple divide-and-conquer implementation of the join-based algorithm for bulk-insert.
Both recursive calls can be executed in parallel.
The join operation used here differs from the version explained in this article, instead join2 is used, which misses the second parameter k.
bulkInsert:
I.sort
bulklInsertRec
bulkInsertRec:
if k = 1:
forall e in I: T.insert
else
m := ⌊size / 2⌋
:= split
bulkInsertRec
|| bulkInsertRec
T ← join2
Execution time
Sorting is not considered in this analysis.
This can be improved by using parallel algorithms for splitting and joining.
In this case the execution time is.
Work

Pipelining

Another method of parallelizing bulk operations is to use a pipelining approach.
This can be done by breaking the task of processing a basic operation up into a sequence of subtasks.
For multiple basic operations the subtasks can be processed in parallel by assigning each subtask to a separate processor.
  1. First the bulk of elements to insert must be sorted.
  2. For each element in the algorithm locates the according insertion position in. This can be done in parallel for each element since won't be mutated in this process. Now must be divided into subsequences according to the insertion position of each element. For example is the subsequence of that contains the elements whose insertion position would be to the left of node.
  3. The middle element of every subsequence will be inserted into as a new node. This can be done in parallel for each since by definition the insertion position of each is unique. If contains elements to the left or to the right of, those will be contained in a new set of subsequences as or.
  4. Now possibly contains up to two consecutive red nodes at the end of the paths form the root to the leaves, which needs to be repaired. Note that, while repairing, the insertion position of elements have to be updated, if the corresponding nodes are affected by rotations.
  5. * If two nodes have different nearest black ancestors, they can be repaired in parallel. Since at most four nodes can have the same nearest black ancestor, the nodes at the lowest level can be repaired in a constant number of parallel steps.
  6. * This step will be applied successively to the black levels above until is fully repaired.
  7. The steps 3 to 5 will be repeated on the new subsequences until is empty. At this point every element has been inserted. Each application of these steps is called a stage. Since the length of the subsequences in is and in every stage the subsequences are being cut in half, the number of stages is.
  8. * Since all stages move up the black levels of the tree, they can be parallelised in a pipeline. Once a stage has finished processing one black level, the next stage is able to move up and continue at that level.
    Execution time
Sorting is not considered in this analysis.
Also, is assumed to be smaller than, otherwise it would be more efficient to construct the resulting tree from scratch.