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Binary heap Created 2026-10-13 Updated 2026-10-24
A binary heap represents a priority queue as a complete binary tree obeying a heap order. We use the minimum convention: the priority of every parent in a rooted tree is no greater than the priorities of its children in a rooted tree. Equal priorities are allowed. Each stored entry may be a record carrying a priority key; the numerical example below displays only keys. Reversing the inequalities gives the maximum convention.
The complete binary tree can be stored directly in a zero-indexed array : read its levels from top to bottom, left to right. The root of a tree is at index . The possible children in a rooted tree of index are at and , and an index denotes a present entry only when it is less than . For , its parent in a rooted tree is at , using the floor function. Thus the links of the complete binary tree are determined by arithmetic rather than separately stored pointers.
ConsiderThe children in a rooted tree of index have keys ; those of index have keys ; those of index have keys . The six comparisons , , , , , establish the heap order. At index , the parent in a rooted tree is index , carrying key .
The array is visibly not a sorted array: precedes , and precedes . Nor is the last entry a maximum: it is , whereas the maximum is at index . The heap order compares ancestors and descendants, without imposing an order between separate branches. Nevertheless, following the graph path from the root of a tree to any entry gives a chain of nondecreasing priorities. This short argument proves that
peek can return , a minimum, without inspecting the other entries.Insertion appends an entry to the array, preserving the shape of the complete binary tree, then uses sift up to repair the heap order. Insert key into the displayed binary heap. It is appended at index , and exchanges with indices , , then :The moving key follows ; the keys each move down one level. Only this chain needs repair.
Now apply
extract-min. Save key , move the last key to the root of a tree, and shorten the array to length . Use sift down, always choosing the smaller of the present children in a rooted tree:First exchanges with key at index , since ; then with key at index , since . At index , the possible children in a rooted tree would be indices , both outside the shortened array, so the process stops. The returned key is , and the remaining binary heap is exactly the original one.
For a nonempty binary heap with entries, the deepest level is , using the floor function and logarithm. Each exchange moves one level and uses only a bounded number of comparisons. The resulting time complexity bounds, expressed in Big O notation, are:
| Operation | Time complexity | Reason |
|---|---|---|
peek | Read the root of a tree. | |
| Insert | for | Append, then sift up through at most the levels of the complete binary tree. |
extract-min | for | Move the last entry, then sift down through at most those levels. |
| Build from an arbitrary array | Apply sift down from the last index with children in a rooted tree back to index . |
These bounds assume constant-cost indexed access, key comparisons and exchanges, and available array capacity for insertion. Allocating and copying a larger array is a separate cost. Empty and one-entry cases take constant work.
To build a binary heap, begin at index and work backwards. Entries without children in a rooted tree already obey the heap order. Each sift down, also called heapify down, starts with every present smaller rooted tree below it already repaired. Most entries are close to the bottom and can move very little; successively longer repair distances occur for geometrically fewer entries. Summing these distances gives work, rather than charging to every entry. MIT's notes give this bottom-up construction and contrast it with repeated insertion.[ref]
https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/40d4851e550507ca14dc778b9b2266cc_MIT6_006S20_lec8.pdf. MIT 6.006, lecture 8: implicit indices, heap repair and linear build, pp. 3–6
The Dijkstra algorithm additionally needs to decrease the priority of an existing vertex when its tentative distance improves. Given its current index, lowering the key followed by sift up takes time for . An indexed binary heap maintains a separate map from item identity to its current index and updates that map after every exchange. For vertices numbered in a fixed range, the map can be another array, giving constant-time index lookup. An ordinary unindexed binary heap supplies no such guarantee: locating an arbitrary item can require scanning entries. Item identity and priority must remain distinct, especially when several vertices have equal tentative distances.
The attractive compromise is local organization with global usefulness. A binary heap maintains only the heap order, yet every
extract-min gives the next smallest key. Repeated extraction of the original example gives , so a priority queue also supplies a sorting algorithm without storing a sorted array between extractions. Binary tree Created 2026-10-13 Updated 2026-10-14
A binary tree, in the data structure convention used here, is a rooted tree in which each vertex has zero, one or two children in a rooted tree, occupying distinct left and right positions. A single left child and a single right child give different binary trees. A vertex with no children in a rooted tree is called a leaf. An empty binary tree has no root of a tree and is also allowed as a storage case. [ref]
For example, give the root of a tree label , its left child label and its right child label . Attach as the left and right children in a rooted tree of , and as those of . This binary tree has seven vertices, six edges, and height of a binary tree : a longest downward graph path contains two edges. The binary search tree lookup illustration below shows this shape alongside the binary tree consisting only of successive right children , whose height of a binary tree is .
Left and right positions specify an ordered shape, not a numerical total order on stored labels. Swapping labels and in the first example still gives a binary tree, but destroys the binary search tree property. Nor must a binary tree be balanced: the right-child chain is just as valid. The same shape convention supports a binary heap, a Huffman code, or a binary search tree, each with its own additional rules. The useful idea is to describe two smaller subtrees at each vertex, making structural recursion natural.
For a precise recursive description, a finite binary tree is either empty or a root of a tree carrying a value together with an ordered pair of disjoint smaller binary trees. The entries of this ordered pair are its left and right subtrees, each of which may be empty. Thus two binary tree child slots exist even when fewer than two children in a rooted tree exist; an empty slot does not denote an extra vertex. This decomposition is the basis of structural recursion.[ref]
4.4 Symbol Tables. Introduction to Programming in Java, section 4.4, Binary search trees: recursive definition and node representation
Left and right binary tree child slots distinguish shapes even with just one child
. Solid circles are vertices; outlined squares mark empty binary tree child slots. The first two binary trees differ only in the occupied slot at . The last two have the same four labels but different height of a binary tree. Details.
In the four-vertex mixed example, has left child and right child , while has only a right child . Count its vertices by structural recursion: the empty binary tree contributes zero and each nonempty binary tree contributes one plus the counts of its left and right subtrees. Thus contributes , and contributes . The same decomposition computes height of a binary tree, supports inorder traversal, and supplies short proofs by structural induction. It separates shape from the values stored in a data structure.
Ignoring values, let count finite binary tree shapes with vertices, retaining left and right slots and counting the empty shape once. Splitting at the root of a tree gives the recurrence relationFor each left subtree size , choose its shape and the right subtree shape independently. This yields the Catalan numbers: , , and . The two shapes at are exactly the left-only and right-only examples. Forgetting the slot distinction would change the counting problem.
Bottom-up heap construction Created 2026-10-24 Updated 2026-10-31
Construct a binary heap from an array by processing internal nodes backwards, applying sift down at each. The complete binary tree shape ensures that the subtrees below the current node have already been repaired. The number of nodes whose height is at least is at most , so charging each repair by its height gives total work at most a constant multiple of . The construction takes time in either heap orientation, whereas inserting entries separately has an worst-case bound.
Complete binary tree Created 2026-10-13 Updated 2026-10-14
A complete binary tree is a binary tree whose levels are full except possibly the deepest, which is filled from left to right with no gaps.[ref]
9.1 Binary trees. Data Structures and Algorithms, section 9.1.1: complete shape and binary heap example
Here a level groups vertices by their graph distance from the root of a tree. A binary heap uses this shape to maintain logarithmic height of a binary tree. Completeness differs from the two-children condition of a full binary tree.
Full binary trees, complete binary trees, and perfect binary trees impose different shape conditions
. Each panel states all three properties. Outlined positions expose gaps in the last level. In particular, the five-vertex example in the lower left is both full and complete but is not perfect. Details.
For a six-vertex example, give two children in a rooted tree , give two children in a rooted tree , and give only a left child . The levels have vertices, and the last occupies its first three positions, so this is a complete binary tree. It is not a full binary tree, since has exactly one child. Moving to the right slot of preserves and height of a binary tree , but leaves a gap before and destroys completeness. Merely keeping the deepest vertices near the left side is insufficient; every earlier position must be occupied.
Number positions by the left-to-right level order of breadth-first search, beginning with root position . The children of position have positions and . A nonempty complete binary tree occupies exactly positions , so its height of a binary tree is : positions form level . This consecutive occupancy explains the compact array representation of a binary heap, while heap order is a separate condition on values. Neither completeness nor the binary tree definition supplies the binary search tree ordering rule.
Data structure Created 2026-10-13 Updated 2026-10-18
A data structure organizes stored information together with operations for accessing and changing it. Choosing a data structure means choosing which questions should be cheap to answer: an array supports access by numbered position, a hash table supports lookup by key, and a priority queue supports selecting an item by priority. The operations and their time complexity matter as much as the stored values.
Suppose three requests arrive as , where the number is a priority and smaller numbers are more urgent. A FIFO queue removes Ada first because she arrived first. A minimum-oriented priority queue removes Bo first because is the least priority. An array can store the same three records in arrival order, but finding the least priority by scanning this representation takes three key inspections. A binary heap organizes these records so the least priority is immediately available. The information is the same; the representation changes the work needed to use it.
This is a useful design principle: preserve just enough organization for the operations the application needs. Maintaining a sorted array gives the whole order, but moving entries to insert a new key can be expensive. A binary heap keeps a weaker heap order that supports fast repeated selection without maintaining a sorted array. A good data structure makes an algorithm's next step easy.
Dijkstra algorithm 2026-10-14
The Dijkstra algorithm solves the single-source shortest path problem in a finite graph with nonnegative edge weights, finding distances to all reachable vertices. Start with tentative distance and elsewhere. Repeatedly choose the unfinalized vertex with least tentative distance, declare that label permanent, and perform edge relaxation on its outgoing edges. For an undirected graph, inspect both directions. A finite tentative distance always records the cost of a discovered graph path, but can improve before finalization.
In our example, finalizing discovers at , at , and at . Choose next: edge relaxation through lowers to and to . Choose : edge relaxation through lowers to . Finally choose . The permanent distances are therefore , , , ; the evolving tentative distances were and .
Tentative distances fall until the Dijkstra algorithm makes each label permanent
. Details. 
Store a shortest-path predecessor whenever edge relaxation improves a label. The final table contains enough information to reconstruct routes, rather than merely their costs:
| Vertex | Distance from | Shortest-path predecessor | Finalization order |
|---|---|---|---|
| None: source | |||
Following shortest-path predecessors backwards gives ; reversing the order gives . If only this destination matters, stop when is finalized. Stopping when is first discovered would incorrectly return . If the smallest remaining tentative distance is , all remaining vertices are unreachable.
Why is the permanent choice safe? Suppose the selected vertex had a cheaper graph path. Let be the first unfinalized vertex on that graph path. Its preceding vertex was finalized, so edge relaxation has already offered a label no larger than that prefix's cost. Nonnegative remaining edge weights make the whole cost at least this prefix's cost; selection gives . A cheaper graph path would contradict these comparisons. Zero weights are allowed; ties can be resolved arbitrarily. This is the reason the label-setting shortest-path algorithm works, rather than a lucky feature of the drawing.
Negative weights break that reasoning. In a directed graph with costing , costing , and costing , the Dijkstra algorithm would finalize at before discovering the cheaper cost . The Bellman-Ford algorithm handles negative edge weights when no relevant negative directed cycle exists.
With an adjacency list and an indexed binary heap implementing a priority queue, the time complexity is for , using Big O notation. Each vertex is extracted at most once, each edge is inspected once per direction, and each successful edge relaxation decreases a heap key in time. This assumes constant-cost arithmetic and comparisons; exact computation with very large weights adds their representation cost. The queue uses space beyond the graph. Simply scanning all remaining vertices gives instead.[ref]
https://algs4.cs.princeton.edu/44sp/DijkstraSP.java.html. Princeton, DijkstraSP implementation
[ref]https://algs4.cs.princeton.edu/code/javadoc/edu/princeton/cs/algs4/IndexMinPQ.html. Princeton, indexed binary heap operations
The 1959 paper presents the shortest path problem as its second problem, after the minimum spanning tree problem, and grows known shortest routes in increasing distance order.[ref]
https://www.cs.yale.edu/homes/lans/readings/routing/dijkstra-routing-1959.pdf. Dijkstra, A note on two problems in connexion with graphs, pp. 269–271
In his later recollection, Dijkstra dated the shortest-route discovery to 1956: a demonstration for the ARMAC computer used a simplified Dutch railway map and printed the route between towns. A practical request became an enduring algorithm because the same distance rule applies to any suitable graph.[ref]https://garfield.library.upenn.edu/classics1983/A1983QA19900001.pdf. Dijkstra's recollection, published in 1983
Heap order Created 2026-10-13 Updated 2026-10-24
The minimum convention for heap order requires every parent in a rooted tree to have priority no greater than each of its children in a rooted tree. The displayed formula uses the array representation of a binary heap. It permits equal priorities and does not require a sorted array; the maximum convention reverses the inequality.
Indexed priority queue 2026-10-31
A priority queue whose entries have stable identifiers and an auxiliary map from each identifier to its current storage location. An indexed binary heap can therefore support membership and locating an entry in constant time, followed by a key decrease in logarithmic time. Every heap swap must update both affected map entries; otherwise later updates can modify the wrong item.
Meldable heap Created 2026-10-20 Updated 2026-10-31
A heap data structure supporting a meld operation that combines two heaps into one while preserving the heap order. The cost of melding distinguishes implementations such as a leftist heap from ordinary binary heaps.
Store the array as a binary heap with min-heap order. Build it bottom-up by sifting down internal nodes in reverse order, rather than inserting all values one at a time. The sum of subtree heights is , so construction costs . Then perform minimum extractions: output the root, move the last remaining entry to the root, and sift it down to restore the heap order.
Each extraction returns the smallest remaining key, so the outputs are the smallest keys in increasing order. For a heap of current size , an extraction takes time. HenceThe tight worst-case bound for this ordinary binary heap implementation isWhen the extraction term is substantial, worst-case extractions follow root-to-leaf paths of logarithmic length; when is small, the linear construction already dominates. The bound stays valid at , since . Building by separate insertions would lose the linear construction bound and is unnecessary.
Store a complete binary tree in an array , with children and parent . The max-heap invariant says every parent's key is at least its children's keys. Consequently the root is a greatest-key item, and the height is .
Here is a pairwise-exchange implementation of
extractMax:extractMax(H, n):
if n == 0: error "empty priority queue"
answer = H[0]
swap(H[0], H[n-1])
n = n - 1
i = 0
while 2*i + 1 < n:
j = 2*i + 1
if j + 1 < n and key(H[j+1]) > key(H[j]):
j = j + 1
if key(H[i]) >= key(H[j]): break
swap(H[i], H[j])
i = j
return (answer, n)Only the root can initially violate heap order. Each heapify down exchange promotes the larger child, making the previous position valid and moving the only possible violation down one level. It either stops at a valid position or reaches a leaf. The removed root is saved as
answer, and reducing the active array length preserves the complete binary tree shape.| Operation | Worst-case time |
|---|---|
Empty initialization, isEmpty, size | |
maximum | |
insert, increaseKey | |
extractMax, deletion by handle | |
| Build from arbitrary items by bottom-up heap construction |
Insertion into the binary heap places an item at the next leaf and moves it upward until heap order holds. Raising a key uses the same upward repair. Arbitrary deletion replaces its position by the final item and repairs upward or downward as necessary. A handle must be updated on exchanges; locating an item by value alone can require linear search. The boxed main bounds are . Dynamic array resizing is usually amortized or avoided by reserving sufficient capacity.
An indexed binary heap builds the initial queue in time;
extract_min and decrease_key each cost , while a position table supports membership in . There are extractions and at most edge inspections, with at most one successful decrease per inspection. Thus the bound is , or for a connected graph with at least two vertices. The singleton case is constant time. An unindexed heap cannot provide decrease-key by vertex without an additional position structure.Use minimum-hop shortest paths by minimizing the lexicographic order pair : total weight first, number of edges second. Extend a pair across an edge of weight by adding , and compare the full pair in both the priority queue and relaxation:
for v in V: key[v] = (infinity,infinity)
key[s] = (0,0)
Q = indexed min-priority-queue containing all vertices, keyed by key
while Q is not empty:
u = extract-min(Q)
if key[u].distance == infinity: break
for each edge (u,v) of weight w with v still in Q:
candidate = (key[u].distance + w, key[u].hops + 1)
if candidate <lex key[v]:
key[v] = candidate
decrease-key(Q,v,candidate)
for v in V: best[v] = key[v].hopsAll weights are positive, so extending a path strictly increases its first component. The usual Dijkstra's algorithm settled-label proof therefore applies to these ordered pairs: no path through a not-yet-settled vertex can later improve an extracted minimum. Among equal minimum weights, the secondary component selects the fewest edges. Merely breaking extraction ties by vertex name would not implement this objective.
With a binary heap and adjacency lists, the bound is , with extra storage. Unreachable vertices have
best = infinity. In the paper's example, the resulting pairs in order are , so the requested edge counts are .Replace breadth-first search by Dijkstra's algorithm from , using the positive edge weights to compute . The strongly connected components, the condensation graph and the reverse-topological order propagation are unchanged, because reachability ignores weights. Only the label assigned to a reachable vertex changes.
With a binary heap, the total time is , including the linear component-processing stages. More specialized priority queues give their corresponding shortest-path bounds. The proof still compares the precomputed distances from ; it must not add the weight of a path from to the candidate, which would optimize a different quantity.
Assume the subtrees rooted at the children in a rooted tree of already obey the max-heap condition. Compare with each present child and choose an index of the largest of these at most three keys. If that index is , stop. Otherwise exchange the two keys, continue at the chosen child, and repeat. This is sift down, with the inequality reversed from a minimum-oriented binary heap.
The exchange repairs the old root position because its new key dominates both child roots; only the subtree receiving the displaced key can still be defective. This recursion invariant proves correctness. Each exchange descends one level and costs constant work, so the worst-case running time is for subtree height , hence . A small key at the root of a suitably ordered heap must descend to a leaf, so the logarithmic worst-case bound is attained.
All edge lengths are positive, so run the Dijkstra algorithm from every source , retaining distances and a shortest-path tree of its reachable component. Each run takes with a binary heap. The following scans after each run use only additional work.
For a directed graph, inspect every closing edge and consider when is reachable. The tree route from to followed by this edge is a closed walk. Every directed cycle supplies one such upper bound by choosing any of its vertices as ; the shortest route to its predecessor is no longer than the corresponding cycle segment. Positivity lets any repeated portion be removed, so the smallest candidate is the shortest directed-cycle length.
For an undirected graph, do not combine an edge with the distance between its endpoints indiscriminately: that can count traversal of one bridge out and back as a false cycle. Instead inspect only edges absent from the current shortest-path tree, with both endpoints reachable, and considerThe two tree routes and this non-tree edge contain a genuine cycle. If they share a prefix, its cost is counted twice, so that cycle is no longer than the candidate. Every candidate is consequently at least the global shortest-cycle length.
To prove equality is attained, take a shortest cycle and a source on it. If either cycle edge incident to is absent from the tree, its candidate uses a shortest route no longer than the rest of , so it is at most the length of . Otherwise both incident edges are in the tree, and their other endpoints lie in different branches below . Along the remaining segment of , some edge crosses between different branches. Such an edge is not a tree edge. The two distances to its endpoints are bounded by the two corresponding, disjoint segments of from ; their sum plus the crossing edge is at most the length of . Thus some candidate is no greater than the optimum, completing the proof.
Return the smallest finite candidate, or report acyclicity if none exists. The total running time is , as required. The ordinary convention is a simple graph; if parallel edges are allowed, track actual edge identities when excluding tree edges, and treat a self-loop as its own cycle rather than discarding it. This is shortest cycle by repeated shortest-path trees.
A minimum-oriented binary heap is an appropriate alternative. Store a complete binary tree in an array and maintain heap order. Insertion appends a value and moves it upward; removing the root replaces it by the last value and moves that value downward. Each path has levels, and equal values can be stored in separate slots. Both operations therefore take time when capacity is available.
Unlike a binary search tree, the heap does not maintain a complete search order, so finding an arbitrary specified value can require inspections. A simple array implementation also has a capacity issue: a fixed array can fill, and resizing a growable array occasionally costs , giving an amortized insertion guarantee unless allocation is otherwise managed. This capacity cost is an implementation tradeoff, not an inherent inability of heaps to represent the priority queue abstraction.
Priority queue 2026-10-13
A priority queue stores items with comparable priority keys and supports insertion, inspection of a best-priority item, and removal of a best-priority item. Here we use the minimum convention: keys have a total order, and smaller means earlier. The operation
peek returns a stored item of minimum priority without removing it; extract-min removes and returns such an item. Both require a nonempty priority queue, or an explicitly specified empty-case result. A maximum-oriented priority queue reverses the choice.For the requests ,
peek returns Bo and leaves all three requests stored. extract-min returns Bo and leaves Ada and Cy; another peek returns Cy. Inserting now makes Dee the next request selected. This changing collection need not be a sorted array: only the next choice must be available.Distinct items may share a priority. Inserting alongside permits either item to be removed first. A priority queue does not inherently promise arrival order on ties. If that policy matters, compare priority first and a unique increasing arrival number second; the resulting total order makes earlier arrivals win equal-priority ties. This is analogous to the motivation for stable sorting, but it is an extra policy for the priority queue.
The interface specifies behavior; the data structure determines costs. A binary heap implements a priority queue with constant-time
peek and logarithmic-time insertion and extract-min, under the cost assumptions below. Scheduling by urgency or by the time of the next event needs exactly this repeated next-item selection. MIT's notes use these applications to motivate the priority queue interface.[ref]https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/40d4851e550507ca14dc778b9b2266cc_MIT6_006S20_lec8.pdf. MIT 6.006, lecture 8: priority queues and binary heaps, pp. 1–2
Shortest cycle by repeated shortest-path trees Created 2026-10-25 Updated 2026-10-31
In a positive-weight directed graph, run the Dijkstra algorithm from each source and minimize over closing edges. In an undirected graph, minimize only over edges absent from the current shortest-path tree. A tree edge would otherwise give a spurious traversal out and back. A non-tree candidate contains a genuine cycle no longer than its score, even when the two tree routes share a prefix.
For the undirected optimum, choose a shortest cycle and a source on it. If an incident cycle edge is absent from the tree, its candidate is bounded by the cycle length. Otherwise its two neighbours lie in distinct tree branches, and some edge on the remaining cycle segment crosses branches. That edge is non-tree; shortest distances to its endpoints are bounded by the two complementary cycle segments, again giving a candidate no greater than the optimum. Thus the minimum candidate equals the shortest-cycle length. Repeated Dijkstra algorithm runs take running time with a binary heap.
Sift down 2026-10-13
To repair a key in a minimum-oriented binary heap, choose the smaller-priority present entry among its children in a rooted tree. Exchange if that entry is smaller than the moving key, and continue at the new index. Stop when there are no children in a rooted tree or the heap order holds. The smaller rooted trees below the starting entry must already obey the heap order; choosing the larger child could leave a violation with the smaller one.