For any node child nodes are located at either 2*i , 2*i +1 So the parent node could be found by taking the floor of the half of child node.
The total time taken will be N times the complexity of adding a single element to the heap .And adding a single element takes logN time,so That is equal to N*logN.
Decreasing a node value may result in violating the min property. As a result be there would be exchange in the value of parent and child which at max goes up to height of the heap.
The function return minimum value in the heap_Array which is equal to the root value of the heap.
Deletion in a min-heap is in O(1) time.
The property of heap that the value of root must be either greater or less than both of its children makes it work like a priority queue.
Union of two trees increase the order of the resultant by atmost value 1.
The main characteristics of a Fibonacci heap is violated since min[H] must contain one with smallest value.
e use a buffer array to store a fixed number of elements when the buffer is full the content of buffer is moved to heap.As a result the complexity is amotized O(1) .
No, The inorder traversal will not give elements in sorted order. As heap is implemented as either min-heap or max-heap ,the root
will be have highest or lowest value than remaining values of the nodes .So this traversal will not give a sorted list.