Recursive lambda expression to traverse a tree in C#
Ok, I found some free time finally.
Here we go:
class TreeNode
{
public string Value { get; set;}
public List<TreeNode> Nodes { get; set;}
public TreeNode()
{
Nodes = new List<TreeNode>();
}
}
Action<TreeNode> traverse = null;
traverse = (n) => { Console.WriteLine(n.Value); n.Nodes.ForEach(traverse);};
var root = new TreeNode { Value = "Root" };
root.Nodes.Add(new TreeNode { Value = "ChildA"} );
root.Nodes[0].Nodes.Add(new TreeNode { Value = "ChildA1" });
root.Nodes[0].Nodes.Add(new TreeNode { Value = "ChildA2" });
root.Nodes.Add(new TreeNode { Value = "ChildB"} );
root.Nodes[1].Nodes.Add(new TreeNode { Value = "ChildB1" });
root.Nodes[1].Nodes.Add(new TreeNode { Value = "ChildB2" });
traverse(root);
A proper solution, and indeed the idiomatic solution in many functional programming languages, would be the use of a fixed-point combinator. In a nutshell: a fixed-point combinator answers the question “how do I define an anonymous function to be recursive?”. But the solution is so nontrivial that whole articles are written to explain them.
A simple, pragmatic alternative is to “go back in time” to the antics of C: declaration before definition. Try the following (the “factorial” function):
Func<int, int> fact = null;
fact = x => (x == 0) ? 1 : x * fact(x - 1);
Works like a charm.
Or, for a pre-order tree traversal on an object of class TreeNode
which implements IEnumerable<TreeNode>
appropriately to go over its children:
Action<TreeNode, Action<TreeNode>> preorderTraverse = null;
preorderTraverse = (node, action) => {
action(node);
foreach (var child in node) preorderTraverse(child, action);
};
A simple alternative is to “go back in time” to the antics of C and C++: declaration before definition. Try the following:
Func<int, int> fact = null; fact = x => (x == 0) ? 1 : x * fact(x - 1);
Works like a charm.
Yes, that does work, with one little caveat. C# has mutable references. So make sure you don't accidentally do something like this:
Func<int, int> fact = null;
fact = x => (x == 0) ? 1 : x * fact(x - 1);
// Make a new reference to the factorial function
Func<int, int> myFact = fact;
// Use the new reference to calculate the factorial of 4
myFact(4); // returns 24
// Modify the old reference
fact = x => x;
// Again, use the new reference to calculate
myFact(4); // returns 12
Of course, this example is a bit contrived, but this could happen when using mutable references. If you use the combinators from aku's links, this won't be possible.