Binary heap

To add an element to a heap, we can perform this algorithm:

1. Add the element to the bottom level of the heap at the leftmost open space.
2. Compare the added element with its parent; if they are in the correct order, stop.
3. If not, swap the element with its parent and return to the previous step.

Steps 2 and 3, which restore the heap property by comparing and possibly swapping a node with its parent, are called the up-heap operation (also known as bubble-up, percolate-up, sift-up, trickle-up, swim-up, heapify-up, or cascade-up).

The number of operations required depends only on the number of levels the new element must rise to satisfy the heap property. Thus, the insertion operation has a worst-case time complexity of O(log n). For a random heap, and for repeated insertions, the insertion operation has an average-case complexity of O(1).[4][5]

As an example of binary heap insertion, say we have a max-heap

and we want to add the number 15 to the heap. We first place the 15 in the position marked by the X. However, the heap property is violated since 15 > 8, so we need to swap the 15 and the 8. So, we have the heap looking as follows after the first swap:

However the heap property is still violated since 15 > 11, so we need to swap again:

which is a valid max-heap. There is no need to check the left child after this final step: at the start, the max-heap was valid, meaning the root was already greater than its left child, so replacing the root with an even greater value will maintain the property that each node is greater than its children (11 > 5; if 15 > 11, and 11 > 5, then 15 > 5, because of the transitive relation).

The procedure for deleting the root from the heap (effectively extracting the maximum element in a max-heap or the minimum element in a min-heap) while retaining the heap property is as follows:

1. Replace the root of the heap with the last element on the last level.
2. Compare the new root with its children; if they are in the correct order, stop.
3. If not, swap the element with one of its children and return to the previous step. (Swap with its smaller child in a min-heap and its larger child in a max-heap.)

Steps 2 and 3, which restore the heap property by comparing and possibly swapping a node with one of its children, are called the down-heap (also known as bubble-down, percolate-down, sift-down, sink-down, trickle down, heapify-down, cascade-down, extract-min or extract-max, or simply heapify) operation.

So, if we have the same max-heap as before

We remove the 11 and replace it with the 4.

Now the heap property is violated since 8 is greater than 4. In this case, swapping the two elements, 4 and 8, is enough to restore the heap property and we need not swap elements further:

The downward-moving node is swapped with the larger of its children in a max-heap (in a min-heap it would be swapped with its smaller child), until it satisfies the heap property in its new position. This functionality is achieved by the Max-Heapify function as defined below in pseudocode for an array-backed heap A of length length(A). A is indexed starting at 1.

// Perform a down-heap or heapify-down operation for a max-heap
// A: an array representing the heap, indexed starting at 1
// i: the index to start at when heapifying down
Max-Heapify(A, i):
    left ← 2×i
    right ← 2×i + 1
    largest ← i
    
    if left ≤ length(A) and A[left] > A[largest] then:
        largest ← left

    if right ≤ length(A) and A[right] > A[largest] then:
        largest ← right
    
    if largest ≠ i then:
        swap A[i] and A[largest]
        Max-Heapify(A, largest)

For the above algorithm to correctly re-heapify the array, no nodes besides the node at index i and its two direct children can violate the heap property. The down-heap operation (without the preceding swap) can also be used to modify the value of the root, even when an element is not being deleted.

In the worst case, the new root has to be swapped with its child on each level until it reaches the bottom level of the heap, meaning that the delete operation has a time complexity relative to the height of the tree, or O(log n).

Inserting an element then extracting from the heap can be done more efficiently than simply calling the insert and extract functions defined above, which would involve both an upheap and downheap operation. Instead, we can do just a downheap operation, as follows:

1. Compare whether the item we're pushing or the peeked top of the heap is greater (assuming a max heap)
2. If the root of the heap is greater:
   1. Replace the root with the new item
   2. Down-heapify starting from the root
3. Else, return the item we're pushing

Python provides such a function for insertion then extraction called "heappushpop", which is paraphrased below.[6][7] The heap array is assumed to have its first element at index 1.

// Push a new item to a (max) heap and then extract the root of the resulting heap. 
// heap: an array representing the heap, indexed at 1
// item: an element to insert
// Returns the greater of the two between item and the root of heap.
Push-Pop(heap: List<T>, item: T) -> T:
    if heap is not empty and heap[1] > item then:  // < if min heap
        swap heap[1] and item
        _downheap(heap starting from index 1)
    return item

A similar function can be defined for popping and then inserting, which in Python is called "heapreplace":

// Extract the root of the heap, and push a new item 
// heap: an array representing the heap, indexed at 1
// item: an element to insert
// Returns the current root of heap
Replace(heap: List<T>, item: T) -> T:
    swap heap[1] and item
    _downheap(heap starting from index 1)
    return item

Finding an arbitrary element takes O(n) time.

Deleting an arbitrary element can be done as follows:

1. Find the index ${\displaystyle i}$ of the element we want to delete
2. Swap this element with the last element
3. Down-heapify or up-heapify to restore the heap property. In a max-heap (min-heap), up-heapify is only required when the new key of element ${\displaystyle i}$ is greater (smaller) than the previous one because only the heap-property of the parent element might be violated. Assuming that the heap-property was valid between element ${\displaystyle i}$ and its children before the element swap, it can't be violated by a now larger (smaller) key value. When the new key is less (greater) than the previous one then only a down-heapify is required because the heap-property might only be violated in the child elements.

The decrease key operation replaces the value of a node with a given value with a lower value, and the increase key operation does the same but with a higher value. This involves finding the node with the given value, changing the value, and then down-heapifying or up-heapifying to restore the heap property.

Decrease key can be done as follows:

1. Find the index of the element we want to modify
2. Decrease the value of the node
3. Down-heapify (assuming a max heap) to restore the heap property

Increase key can be done as follows:

1. Find the index of the element we want to modify
2. Increase the value of the node
3. Up-heapify (assuming a max heap) to restore the heap property

For a general node located at index i (beginning from 0), we will first derive the index of its right child, ${\displaystyle {\text{right}}=2i+2}$.

Let node i be located in level L, and note that any level l contains exactly ${\displaystyle 2^{l}}$ nodes. Furthermore, there are exactly ${\displaystyle 2^{l+1}-1}$ nodes contained in the layers up to and including layer l (think of binary arithmetic; 0111...111 = 1000...000 - 1). Because the root is stored at 0, the kth node will be stored at index ${\displaystyle (k-1)}$. Putting these observations together yields the following expression for the index of the last node in layer l.

${\displaystyle {\text{last}}(l)=(2^{l+1}-1)-1=2^{l+1}-2}$

Let there be j nodes after node i in layer L, such that

${\displaystyle {\begin{alignedat}{2}i=&\quad {\text{last}}(L)-j\\=&\quad (2^{L+1}-2)-j\\\end{alignedat}}}$

Each of these j nodes must have exactly 2 children, so there must be ${\displaystyle 2j}$ nodes separating i's right child from the end of its layer (${\displaystyle L+1}$).

${\displaystyle {\begin{alignedat}{2}{\text{right}}=&\quad {\text{last(L + 1)}}-2j\\=&\quad (2^{L+2}-2)-2j\\=&\quad 2(2^{L+1}-2-j)+2\\=&\quad 2i+2\end{alignedat}}}$

As required.

Noting that the left child of any node is always 1 place before its right child, we get ${\displaystyle {\text{left}}=2i+1}$.

If the root is located at index 1 instead of 0, the last node in each level is instead at index ${\displaystyle 2^{l+1}-1}$. Using this throughout yields ${\displaystyle {\text{left}}=2i}$ and ${\displaystyle {\text{right}}=2i+1}$ for heaps with their root at 1.

Every node is either the left or right child of its parent, so we know that either of the following is true.

1. ${\displaystyle i=2\times ({\text{parent}})+1}$
2. ${\displaystyle i=2\times ({\text{parent}})+2}$

Hence,

${\displaystyle {\text{parent}}={\frac {i-1}{2}}\;{\textrm {or}}\;{\frac {i-2}{2}}}$

Now consider the expression ${\displaystyle \left\lfloor {\dfrac {i-1}{2}}\right\rfloor }$.

If node ${\displaystyle i}$ is a left child, this gives the result immediately, however, it also gives the correct result if node ${\displaystyle i}$ is a right child. In this case, ${\displaystyle (i-2)}$ must be even, and hence ${\displaystyle (i-1)}$ must be odd.

${\displaystyle {\begin{alignedat}{2}\left\lfloor {\dfrac {i-1}{2}}\right\rfloor =&\quad \left\lfloor {\dfrac {i-2}{2}}+{\dfrac {1}{2}}\right\rfloor \\=&\quad {\frac {i-2}{2}}\\=&\quad {\text{parent}}\end{alignedat}}}$

Therefore, irrespective of whether a node is a left or right child, its parent can be found by the expression:

${\displaystyle {\text{parent}}=\left\lfloor {\dfrac {i-1}{2}}\right\rfloor }$