the expression for the running time for the insert operation follows directly from that of the merge operation that isthe time required for the insert operation in the worst case is notes ( ) (isgto( )where is the null path length of the heap into which the item is inserted if you assume that two keys can be compared in constant timethe running time for insert becomes simply (log )where is the number of nodes in the tree into which the item is inserted task discuss the use of enqueue method removing items from leftist heap the findmin method locates the item with the smallest key in given priority queue and the dequeuemin method removes it from the queue since the smallest item in heap is found at the rootthe findmin operation is easy to implement program shows how it can be done clearlythe running time of the findmin operation is ( public class leftistheap extends binarytree implements mergeablepriorityqueue protected int nullpathlengthpublic comparable findmin (if (isempty ()throw new containeremptyexeception ()return (comparablegetkey ()/program leftistheap class findmin method since the smallest item in heap is at the rootthe dequeuemin operation must delete the root node since leftist heap is binary heapthe root has at most two children in general when the root is deletedyou are left with two non-empty leftist heaps since you already have an efficient way to merge leftist heapsthe solution is to simply merge the two children of the root to obtain single heap againprogram shows how the dequeuemin operation of the leftistheap class can be implemented public class leflistheap extends binarytree implements mergeablepriorityqueue protected int nullpathlengthpublic comparable dequeuemin (if (isempty ()lovely professional university