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 (re-)balancing is not perfect, but guarantees searching in O(\log n) time, where n is the number of entries in the tree. The insert and delete operations, along with tree rearrangement and recoloring, also execute in O(\log n) time. Tracking the color of each node requires only one bit of information per node because there are only two colors (due to memory alignment present in some programming languages, the real memory consumption may differ). 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 (uncolored) 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 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 4 unbalanced cases and one default balanced case. The original algorithm used 8 unbalanced cases, but reduced that to 6 unbalanced cases. Sedgewick showed that the insert operation can be implemented in just 46 lines of Java code. 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

A red–black tree is a special type of binary search tree, used in computer science to organize pieces of comparable data, such as text fragments or numbers (as e.g. the numbers in figures 1 and 2). The nodes carrying keys and/or data are frequently called "internal nodes", but to make this very specific they are also called non-NIL nodes in this article. The leaf nodes of red–black trees (   NIL   in figure 1) do not contain keys or data. These "leaves" need not be explicit individuals in computer memory: a NULL pointer can—as in all binary tree data structures— encode the fact that there is no child node at this position in the (parent) node. Nevertheless, by their position in the tree, these objects are in relation to other nodes that is relevant to the RB-structure, it may have parent, sibling (i.e., the other child of the parent), uncle, even nephew node; and may be child—but never parent—of another node. It is not really necessary to attribute a "color" to these end-of-path objects, because the condition "is or " is implied by the condition "is " (see also this remark). Figure 2 shows the conceptually same red–black tree without these NIL leaves. To arrive at the same notion of a path, one must notice that e.g., 3 paths run through the node 1, namely a path through 1left plus 2 added paths through 1right, namely the paths through 6left and 6right. This way, these ends of the paths are also docking points for new nodes to be inserted, fully equivalent to the NIL leaves of figure 1. Instead, to save a marginal amount of execution time, these (possibly many) NIL leaves may be implemented as pointers to one unique (and black) sentinel node (instead of pointers of value NULL). As a conclusion, the fact that a child does not exist (is not a true node, does not contain data) can in all occurrences be specified by the very same NULL pointer or as the very same pointer to a sentinel node. Throughout this article, either choice is called NIL node and has the constant value. The black depth of a node is defined as the number of black nodes from the root to that node (i.e. the number of black ancestors). 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 (alternatively, it could be defined as the black depth of any leaf node). The black height of a node is the black height of the subtree rooted by it. In this article, the black height of a NIL node shall be set to 0, because its subtree is empty as suggested by figure 2, 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 red–black tree: 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 h 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 n of entries, i.e., which is not the case for ordinary binary search trees. For a mathematical proof see section Proof of bounds. Red–black trees, like all binary search trees, allow quite efficient sequential access (e.g. in-order traversal, that is: in the order Left–Root–Right) of their elements. But they support also asymptotically optimal direct access via a traversal from root to leaf, resulting in O(\log n) 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 3. 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 O(\log n) 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 O(\log n) 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 O(m) to O(\log m) where m is the number of elements with colliding hashcodes.

Operations

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. It requires in the worst case a small number, O(\log n) in Big O notation, where n is the number of objects in the tree, on average or amortized O(1), a constant number, of color changes (which are very quick in practice); and no more than three tree rotations (two for insertion). If the example implementation below is not suitable, other implementations with explanations may be found in Ben Pfaff’s annotated C library GNU libavl (v2.0.3 as of June 2019). The details of the insert and removal operations will be demonstrated with example C++ code, which uses the type definitions, macros below, and the helper function for rotations:

Notes to the sample code and diagrams of insertion and removal

The proposal breaks down both, insertion and removal (not mentioning some very simple cases), into six constellations of nodes, edges and colors, which are called cases. The proposal contains for both, insertion and removal, exactly one case that advances one black level closer to the root and loops, the other five cases rebalance the tree of their own. The more complicated cases are pictured in a diagram.

Insertion

Insertion begins by placing the new (non-NIL) node, say N, at the position in the binary search tree of a NIL 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. (Frequently, this positioning is the result of a search within the tree immediately preceding the insert operation and consists of a node together with a direction with .) The newly inserted node is temporarily colored red so that all paths contain the same number of black nodes as before. But if its parent, say P, is also red then this action introduces a red-violation. The rebalancing loop of the insert operation has the following invariant:

Notes to the insert diagrams

Insert case I1

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

Insert case I2

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 (= 2 tree levels) higher.

Insert case I3

Insert case I2 has been executed for times and the total height of the tree has increased by 1, now being h~. The current node N is the (red) root of the tree, and all RB-properties are satisfied.

Insert case I4

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 I5

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 (i.e., if N is the left child of the right child of G or the right child of the left child of G). A -rotation at P switches the roles of the current node N and its parent P. The rotation adds paths through N (those in the subtree labeled 2, see diagram) and removes paths through P (those in the subtree labeled 4). But both P and N are red, so requirement 4 is preserved. Requirement 3 is restored in case 6.

Insert case I6

The current node N is now certain to be an "outer" grandchild of G (left of left child or right of right child). Now -rotate 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 (non-NIL), then we can swap its value with its in-order successor (the leftmost child of the right subtree), and then delete the successor instead. Since the successor is leftmost, it can only have a right child (non-NIL) or no child at all.
• When the deleted node has only 1 child (non-NIL). In this case, just replace the node with its child, and color it black. The single child (non-NIL) 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 (both NIL) and is the root, replace it with NIL. The tree is empty.
• When the deleted node has no children (both NIL), and is red, simply remove the leaf node.
• When the deleted node has no children (both NIL), and is black, deleting it will create an imbalance, and requires a fixup, 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 (⇔ only NIL children). In the first iteration, N is replaced by NIL. The rebalancing loop of the delete operation has the following invariant:

Notes to the delete diagrams

Delete case D1

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 D2

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 (= 1 tree level) higher.

Delete case D3

The sibling S is red, so P and the nephews C and D have to be black. A -rotation 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 D4, D5, or D6 are able to restore the RB-shape.

Delete case D4

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 D5

The sibling S is black, S’s close child C is red, and S’s distant child D is black. After a -rotation 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 ( in the diagram).

Delete case D6

The sibling S is black, S’s distant child D is red. After a -rotation 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 ( in the diagram), 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 (i.o.w. passing through D and node 3 in the diagram) 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 h\in\N there is a red–black tree of height h with nodes ( is the floor function) and there is no red–black tree of this tree height with fewer nodes—therefore it is minimal. Its black height is     (with black root) or for odd h (then with a red root) also   (h-1)/2~. 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 h=1 with red root is minimal. This is in agreement with A minimal RB tree (RBh in figure 4) of height h>1 has a root whose two child subtrees are of different height. The higher child subtree is also a minimal RB tree, RBh–1, containing also a longest path that defines its height h!!-!!1; it has m_{h-1} nodes and the black height The other subtree is a perfect binary tree of (black) height s having black nodes—and no red node. Then the number of nodes is by induction The graph of the function m_h is convex and piecewise linear with breakpoints at where k \in \N. The function has been tabulated as m_h= A027383(h–1) for h\geq 1. The inequality 9>8=2^3 leads to 3 > 2^{3/2}, which for odd h leads to So in both, the even and the odd case, h is in the interval with n being the number of nodes. A red–black tree with n nodes (keys) 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 red–black trees: 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. t1 and t2 and a key k, where t1 < k < t2 , i.e. all keys in t1 are less than k, and all keys in t2 are greater than k. It returns a tree containing all elements in t1 , t2 also as k. t1 , root k and right subtree t2 . If both t1 and t2 have black root, set k to be red. Otherwise k is set black. t1 has larger black height than t2 (the other case is symmetric). Join follows the right spine of t1 until a black node c, which is balanced with t2 . At this point a new node with left child c, root k (set to be red) and right child t2 is created to replace c. The new node may invalidate the red–black invariant because at most three red nodes can appear in a row. This can be fixed with a double rotation. If double red issue propagates to the root, the root is then set to be black, restoring the properties. The cost of this function is the difference of the black heights between the two input trees. The join algorithm is as follows: function joinRightRB(TL, k, TR): if (TL.color=black) and (TL.blackHeight=TR.blackHeight): return Node(TL,⟨k,red⟩,TR) T'=Node(TL.left,⟨TL.key,TL.color⟩,joinRightRB(TL.right,k,TR)) if (TL.color=black) and (T'.right.color=T'.right.right.color=red): T'.right.right.color=black; return rotateLeft(T') return T' /* T[recte T'] / function joinLeftRB(TL, k, TR): / symmetric to joinRightRB / function join(TL, k, TR): if TL.blackHeight>TR.blackHeight: T'=joinRightRB(TL,k,TR) if (T'.color=red) and (T'.right.color=red): T'.color=black return T' if TR.blackHeight>TL.blackHeight: / symmetric */ if (TL.color=black) and (TR.color=black): return Node(TL,⟨k,red⟩,TR) return Node(TL,⟨k,black⟩,TR) The split algorithm is as follows: function split(T, k): if (T = nil) return (nil, false, nil) if (k = T.key) return (T.left, true, T.right) if (k < T.key): (L',b,R') = split(T.left, k) return (L',b,join(R',T.key,T.right)) (L',b,R') = split(T.right, k) return (join(T.left,T.key,L'),b,T.right) The union of two red–black trees t1 and t2 representing sets A and B, is a red–black tree t that represents A ∪ B . The following recursive function computes this union: function union(t1, t2): if t1 = nil return t2 if t2 = nil return t1 (L1,b,R1)=split(t1,t2.key) proc1=start: TL=union(L1,t2.left) proc2=start: TR=union(R1,t2.right) wait all proc1,proc2 return join(TL, t2.key, TR) Here, split is presumed to return two trees: one holding the keys less its input key, one holding the greater keys. (The algorithm is non-destructive, but an in-place destructive version exists also.) 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 m and n(\ge m). 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 m=1, the join-based implementation has the same computational directed acyclic graph (DAG) 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 n of items where n\to\infty. 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 I into a tree T.

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 I and T in multiple parts and perform the insertions on these parts in parallel. Note that in Step 3 the constraints for splitting I 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(T, I, k): I.sort bulklInsertRec(T, I, k) bulkInsertRec(T, I, k): if k = 1: forall e in I: T.insert(e) else m := ⌊size(I) / 2⌋ (T1, _, T2) := split(T, I[m]) bulkInsertRec(T1, I[0 .. m], ⌈k / 2⌉) || bulkInsertRec(T2, I[m + 1 .. size(I) - 1], ⌊k / 2⌋) T ← join2(T1, T2)

Execution time

Sorting I 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.

Execution time

Sorting I is not considered in this analysis. Also, |I| is assumed to be smaller than |T|, otherwise it would be more efficient to construct the resulting tree from scratch.

Work

References and notes