Inserting an element into a stack (push operation) is done in constant time, hence O(1).

Inserting an element into an AVL tree takes O(log n) time due to the need to maintain balance after insertion.

Level-order traversal visits each node once, resulting in a time complexity of O(n).

Linear search checks each element one by one, resulting in a time complexity of O(n).

Linear search checks each element one by one, resulting in a time complexity of O(n).

Merge sort divides the array and merges sorted halves, resulting in a time complexity of O(n log n).

Merging two binary trees involves visiting each node, resulting in a time complexity of O(n), where n is the total number of nodes in both trees.

Merging two sorted arrays takes linear time, O(m + n), as we traverse both arrays.

Merging two sorted arrays can be done in linear time, O(n), where n is the total number of elements in both arrays.

Merging two sorted binary trees involves visiting each node, resulting in a time complexity of O(n), where n is the total number of nodes.

Merging two sorted linked lists can be done in linear time, O(n), where n is the total number of elements in both lists.

Binary search has a time complexity of O(log n) because it repeatedly divides the search interval in half.

The time complexity of performing a breadth-first search (BFS) on a graph is O(V + E), where V is the number of vertices and E is the number of edges.

A level-order traversal of a Red-Black Tree visits each node once, resulting in a time complexity of O(n).

Popping an element from a queue implemented using a linked list can be done in constant time, O(1), if we maintain a pointer to the front of the queue.

Popping an element from a stack is also done in constant time, O(1), as it involves removing the top element.

Post-order traversal visits each node exactly once, resulting in a time complexity of O(n), where n is the number of nodes in the tree.

Prim's algorithm has a time complexity of O(V^2) when using an adjacency matrix.

Pushing an element onto a stack implemented using a linked list takes constant time, O(1), because it involves adding a new node at the head of the list.

Pushing an element onto a stack is done in constant time, hence O(1).

Quicksort has an average-case time complexity of O(n log n) due to the divide-and-conquer approach.

In the best case, quicksort divides the array into two equal halves, leading to a time complexity of O(n log n).

Removing an element from a queue is O(1) because it involves removing the front element.

Reversing a linked list requires visiting each node once, resulting in O(n) time complexity.

Reversing a queue using a stack involves transferring all elements to the stack and then back to the queue, resulting in a time complexity of O(n).

Reversing a stack using another stack takes O(n) time, as each element is pushed and popped once.

Reversing a stack using recursion has a time complexity of O(n) as each element is processed once.

The time complexity of searching for a value in a balanced binary search tree is O(log n).

In a balanced binary search tree, the average time complexity for searching is O(log n), but in the worst case (unbalanced), it can be O(n).

Searching for a value in a Red-Black tree takes O(log n) time due to its balanced structure.