public void Invert_SimpleTree_ShouldReturnInvertedGraph()
{
// arrange
var target = new GraphInvertor();
var root = new Node
{
Value = "1",
Childs = new List <Node>
{
new Node("2")
}
};
// act
var result = target.Invert(root);
// assert
var oldRoot = result.Single(n => n.Value == "1");
oldRoot.Childs.Should().BeEmpty();
var secondNode = result.Single(n => n.Value == "2");
secondNode.Childs.Should().HaveCount(1).And.Contain(oldRoot);
}
public void Invert_ShouldReturnInvertedGraph()
{
// arrange
var target = new GraphInvertor();
var root = new Node
{
Value = "1",
Childs = new List <Node>
{
new Node {
Value = "2", Childs = new List <Node> {
new Node("3"), new Node("4")
}
},
new Node {
Value = "5", Childs = new List <Node> {
new Node("6"), new Node("7")
}
}
}
};
// act
var result = target.Invert(root);
// assert
var oldRoot = result.Single(n => n.Value == "1");
oldRoot.Childs.Should().BeEmpty();
var secondNode = result.Single(n => n.Value == "2");
secondNode.Childs.Should().HaveCount(1).And.Contain(oldRoot);
var fifthNode = result.Single(n => n.Value == "5");
fifthNode.Childs.Should().HaveCount(1).And.Contain(oldRoot);
// old leafs
var thirdNode = result.Single(n => n.Value == "3");
thirdNode.Childs.Should().HaveCount(1).And.Contain(secondNode);
var forthNode = result.Single(n => n.Value == "4");
forthNode.Childs.Should().HaveCount(1).And.Contain(secondNode);
var sixthNode = result.Single(n => n.Value == "6");
sixthNode.Childs.Should().HaveCount(1).And.Contain(fifthNode);
var seventhNode = result.Single(n => n.Value == "7");
seventhNode.Childs.Should().HaveCount(1).And.Contain(fifthNode);
}