Question

Give the psuedo code for inserting and deleting an element in a binary min heap and binary max heap data structure. Discuss the worst case running time for each pseudo code.

Answer 1

--max-heap insertion
Step 1 − Create a new node at the end of heap.
Step 2 − Assign new value to the node.
Step 3 − Compare the value of this child node with its parent.
Step 4 − If value of parent is less than child, then swap them.
Step 5 − Repeat step 3 & 4 until Heap property holds.

--max-heap deletion
Step 1 − Remove root node.
Step 2 − Move the last element of last level to root.
Step 3 − Compare the value of this child node with its parent.
Step 4 − If value of parent is less than child, then swap them.
Step 5 − Repeat step 3 & 4 until Heap property holds.

-- min -heap insertion
Step 1- We add a new key at the end of the tree.
Step 2- IF new key is greater than its parent, then we don’t need to do anything.
Step 3- Otherwise, we need to traverse up to fix the violated heap property.

-- min-heap deletion
Step 1- we shall delete the minimum element from min heap, ie, we shall always delete the root node in each deletion operation, and place the last node in root node position.
Step 2- Since we are placing a leaf node in root node, its guaranteed that heap property shall be violated.
Step 3- We shall then call a special function called heapify() function recursively to make sure that heap property is satisfied.

void heapify(minHeap *hp, int i) {
int smallest = (LCHILD(i) < hp->size && hp->elem[LCHILD(i)].data < hp->elem[i].data) ? LCHILD(i) : i ;
if(RCHILD(i) < hp->size && hp->elem[RCHILD(i)].data < hp->elem[largest].data) {
smallest = RCHILD(i) ;
}
if(smallest != i) {
swap(&(hp->elem[i]), &(hp->elem[smallest])) ;
heapify(hp, smallest) ;
}
}

Complexity:  at worst case, the new node may be lower than all nodes in the heap, that it needs to be compared with log(n) elements at most to be put into the root node position. So insertion operation is normally considered as an O(logn) operation.

Running time for deletion is O(logn).