G is a tree, per node has a special path from the root. So, both BFS and DFS have the exact tree, and the tree is the exact as G.
<h3>What are DFS and BFS?</h3>
An algorithm for navigating or examining tree or graph data structures is called depth-first search. The algorithm moves as far as it can along each branch before turning around, starting at the root node.
The breadth-first search strategy can be used to look for a node in a tree data structure that has a specific property. Before moving on to the nodes at the next depth level, it begins at the root of the tree and investigates every node there.
First, we reveal that G exists a tree when both BFS-tree and DFS-tree are exact.
If G and T are not exact, then there should exist a border e(u, v) in G, that does not belong to T.
In such a case:
- in the DFS tree, one of u or v, should be a prototype of the other.
- in the BFS tree, u and v can differ by only one level.
Since, both DFS-tree and BFS-tree are the very tree T,
it follows that one of u and v should be a prototype of the other and they can discuss by only one party.
This means that the border joining them must be in T.
So, there can not be any limits in G which are not in T.
In the two-part of evidence:
Since G is a tree, per node has a special path from the root. So, both BFS and DFS have the exact tree, and the tree is the exact as G.
The complete question is:
We have a connected graph G = (V, E), and a specific vertex u ∈ V.
Suppose we compute a depth-first search tree rooted at u, and obtain a tree T that includes all nodes of G.
Suppose we then compute a breadth-first search tree rooted at u, and obtain the same tree T.
Prove that G = T. (In other words, if T is both a depth-first search tree and a breadth-first search tree rooted at u, then G cannot contain any edges that do not belong to T.)
To learn more about DFS and BFS, refer to:
brainly.com/question/13014003
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