When analyzing the time complexity of an algorithm, it is important to understand how the input size affects the running time. In the case of breadth-first traversal, also known as breadth-first search (BFS), we will examine how the time complexity depends on the number of nodes in the graph.

BFS is a graph traversal algorithm that explores all the vertices of a graph in breadth-first fashion, i.e., it visits all the nodes at the same level before moving on to the next level. This algorithm is commonly used to find the shortest path between two nodes in an unweighted graph.

Let’s consider a graph with V vertices and E edges. To perform a breadth-first traversal, we typically use a queue data structure to keep track of the nodes that need to be visited. The algorithm starts by enqueuing the starting node and then repeatedly dequeues a node, visits its neighbors, and enqueues them if they have not been visited yet.

The time complexity of BFS can be analyzed as follows:

• Enqueuing the starting node takes O(1) time.
• For each vertex, we visit all of its neighbors exactly once. This requires traversing all the edges connected to that vertex. Since we have V vertices and E edges, the total number of edges traversed is at most E.

Therefore, the overall time complexity of BFS can be expressed as O(V + E).

It is important to note that the time complexity of BFS may vary depending on the implementation. For example, if we use an adjacency matrix to represent the graph, the time complexity for visiting neighbors can be O(V), resulting in a time complexity of O(V^2).

In conclusion, the time complexity of breadth-first traversal is typically O(V + E), where V is the number of vertices and E is the number of edges in the graph. It is a relatively efficient algorithm for exploring a graph in a breadth-first manner, especially for sparse graphs where the number of edges is relatively small compared to the number of vertices.

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