I am trying to solve a problem which is: Find all ancestors of a particular node in a binary tree.
Input: root, targetNode
Output: An array/list containing the ancestors
Suppose, we have the above binary tree as an example. We want to find the ancestors of the node 4. The output should be [3, 5, 2, 4]. If the node is 8, the output is [3, 1, 8]
To solve this, I have written a function which implements DFS.
var ancestor = function(root, target) {
var isFound = false;
const dfs = (node, curr) => {
if (node === null) {
return curr;
}
if (node.val === target.val) {
curr.push(node.val);
isFound = true;
return curr;
}
curr.push(node.val);
const left = dfs(node.left, curr);
if (!isFound) {
const right = dfs(node.right, curr);
curr.pop();
return right;
} else {
curr.pop();
return left;
}
}
console.log(dfs(root, []));
};
But it is not returning the correct ouput. For example, if the targetNode is 7, the output is [3], if the targetNode is 8, the output is also [3]. If I remove the line curr.pop() the output is also invalid. for targetNode 7 it is [3 , 5, 6, 2, 7]. I think I found the issue where I am making mistake. While backtracking, I am doing something wrong with the remove of the node that was pushed in the curr array. If I pass a string instead of the array, it prints the output correctly.
var ancestor = function(root, target) {
var isFound = false;
const dfs = (node, curr) => {
if (node === null) {
return curr;
}
if (node.val === target.val) {
curr += node.val;
isFound = true;
return curr;
}
const left = dfs(node.left, curr + node.val + '->);
if (!isFound) {
const right = dfs(node.right, curr + node.val + '->);
return right;
} else {
return left;
}
}
console.log(dfs(root, ''));
The above code with string instead of array prints the output correctly, If I pass targetNode 7, output is 3->5->2->7
My question is, how to properly unchoose/backtrack here? Or is there anything else that I am doing incorrectly? Thanks in advance.
recursion in its natural setting
Recursion is a functional heritage and so using it with functional style yields the best results. This means avoiding imperative things such as mutations like push and cur += node.val, variable reassignments like isFound = true, and other side effects. We can write ancestor as a simple generator-based function that prepends each node to the output of recursive sub-problem -
const empty =
Symbol("tree.empty")
function node(val, left = empty, right = empty) {
return { val, left, right }
}
function* ancestor(t, q) {
if (t == empty) return
if (t.val == q) yield [t.val]
for (const l of ancestor(t.left, q)) yield [t.val, ...l]
for (const r of ancestor(t.right, q)) yield [t.val, ...r]
}
const mytree =
node(3, node(5, node(6), node(2, node(7), node(4))), node(1, node(0), node(8)))
for (const path of ancestor(mytree, 7))
console.log(path.join("->"))
3->5->2->7
use modules
To finish, I would recommend a module-based approach for this code -
// tree.js
const empty =
Symbol("tree.empty")
function node(val, left = empty, right = empty) {
return { val, left, right }
}
function* ancestor(t, q) {
if (t == empty) return
if (t.val == q) yield [t.val]
for (const l of ancestor(t.left, q)) yield [t.val, ...l]
for (const r of ancestor(t.right, q)) yield [t.val, ...r]
}
function insert(t, val) {
// ...
}
function remove(t, val) {
// ...
}
function fromArray(a) {
// ...
}
// other tree functions...
export { empty, node, ancestor, insert, remove, fromArray }
// main.js
import { node, ancestor } from "./tree.js"
const mytree =
node(3, node(5, node(6), node(2, node(7), node(4))), node(1, node(0), node(8)))
for (const path of ancestor(mytree, 7))
console.log(path.join("->"))
3->5->2->7
private generator
In the previous implementation, our module exposes a generator for ancestor's public interface. Another option is to return undefined when a node cannot be found and has no ancestry. Consider this alternate implementation which hides the generator and requires the caller to null-check the result instead -
const empty =
Symbol("tree.empty")
function node(val, left = empty, right = empty) {
return { val, left, right }
}
function ancestor(t, q) {
function* dfs(t) {
if (t == empty) return
if (t.val == q) yield [t.val]
for (const l of dfs(t.left)) yield [t.val, ...l]
for (const r of dfs(t.right)) yield [t.val, ...r]
}
return Array.from(dfs(t))[0]
}
const mytree =
node(3, node(5, node(6), node(2, node(7), node(4))), node(1, node(0), node(8)))
const result =
ancestor(mytree, 7)
if (result)
console.log(result.join("->"))
else
console.log("no result")
3->5->2->7
You need to check whether the DFS of the right child has found the node.
fix:
const left = dfs(node.left, curr);
if (!isFound) {
const right = dfs(node.right, curr);
if(isFound) {
return right;
}
curr.pop();
return; // return nothing, backtracking
}
return left;
In the array example, your loop iterates through the nodes in a DFS manner, so each of those nodes are connected in that manner. If we count the tree nodes in DFS algorithm, [3 , 5, 6, 2, 7] are actually in order 1, 2, 3, 4 and 5. In this manner, your whole tree in an array should be looking like this; [3 , 5, 6, 2, 7, 4, 1, 0, 8].
So when you find the target value, you pop from the current node and trace it all back up to the head node in DFS algorithm.
I'd either suggest finding a way to get around that, or you could save each of these node's parents. Meaning you could use tuples instead of int arrays (if that's acceptable). An index could look like this = (node value, parent value)
[(3,NULL),(5,3),(6,5),(2,5)...]
And then traceback accordingly...
Im making this anonymus function, and i need it to call itself. Is there any way to do it? I tried the code below, which didnt work...
val example:Char = fun () : Char {
//Some code
if(condition) {
return this();
}
}
What should i replace 'this()' with?
Im pretty new to kotlin, so it would be really helpful with a response
You can't name anonymous functions (either with this syntax, or as a lambda) in Kotlin, and therefore you can't make them recursive either, because you have know way to reference themselves.
If you need recursion, you'll have to create a regular function, and call that:
fun helper() : Char {
if (condition) {
return helper();
}
...
}
val example = helper()
The good news is that you can basically create a regular, named function in any scope. They can be top level outside classes, class members, or just local functions nested within other functions. Wherever you can write down val example = ..., you can also create a function.
Calling an anonymous function sound complicated as there is no name to call it with :)
As I'm learning Kotlin myself at the moment, I tried something and came up with this, hope it helps:
import kotlin.test.Test
import kotlin.test.assertEquals
class StackOverflow51233329 {
#Test
fun test() {
var letter = 'A'
lateinit var example: () -> Char
example = {
letter++
if (letter >= 'C') letter else example()
}
assertEquals('C', example())
}
}
If you want to avoid using lateinit, you could use the Y combinator, which can be used to enable recursion when recursion is impossible directly. Declare this globally:
class RecursiveFunc<T, R>(val f: (RecursiveFunc<T, R>) -> (T) -> R)
fun <T, R> y(f: ((T) -> R) -> (T) -> R): (T) -> R {
val rec = RecursiveFunc<T, R> { r -> f { r.f(r)(it) } }
return rec.f(rec)
}
This code was taken from Rosetta Code. You use it like this:
val fac = y { f: ((Int) -> Int) ->
{ n: Int ->
if (n <= 1) 1 else n * f(n - 1)
}
}
println(fac(10))
f is the recursive function here, with a signature of (Int) -> Int. The rest of the function is pretty much the same as usual, but in lambda form. You can even use the usual function syntax if that's more familiar:
val fac = y { f: (Int) -> Int ->
fun(n: Int): Int {
return if (n <= 1) 1 else n * f(n - 1)
}
}
I know the Function class can be passed as a parameter to another function, like this:
void doSomething(Function f) {
f(123);
}
But is there a way to constrain the arguments and the return type of the function parameter?
For instance, in this case f is being invoked directly on an integer, but what if it was a function accepting a different type?
I tried passing it as a Function<Integer>, but Function is not a parametric type.
Is there any other way to specify the signature of the function being passed as a parameter?
Dart v1.23 added a new syntax for writing function types which also works in-line.
void doSomething(Function(int) f) {
f(123);
}
It has the advantage over the function-parameter syntax that you can also use it for variables or anywhere else you want to write a type.
void doSomething(Function(int) f) {
Function(int) g = f;
g(123);
}
var x = <int Function(int)>[];
int Function(int) returnsAFunction() => (int x) => x + 1;
int Function(int) Function() functionValue = returnsAFunction;
To strongly type a function in dart do the following:
Write down the Function keyword
Function
Prefix it with its return type (for example void)
void Function
Append it with parentheses
void Function()
Put comma separated arguments inside the parentheses
void Function(int, int)
Optionally - give names to your arguments
void Function(int foo, int bar)
Real life example:
void doSomething(void Function(int arg) f) {
f(123);
}
Edit: Note that this answer contains outdated information. See Irn's answer for more up-to-date information.
Just to expand on Randal's answer, your code might look something like:
typedef void IntegerArgument(int x);
void doSomething(IntegerArgument f) {
f(123);
}
Function<int> seems like it would be a nice idea but the problem is that we might want to specify return type as well as the type of an arbitrary number of arguments.
For reference.
int execute(int func(int a, int b)) => func(4, 3);
print(execute((a, b) => a + b));
You can have a function typed parameter or use a typedef
void main() {
doSomething(xToString);
doSomething2(xToString);
}
String xToString(int s) => 's';
typedef String XToStringFn(int s);
void doSomething(String f(int s)) {
print('value: ${f(123)}');
}
void doSomething2(XToStringFn f) {
print('value: ${f(123)}');
}
DartPad example
This is what typedefs are for!
Here is what I've done so far:
struct rep_list {
struct node *head;
struct node *tail;
}
typedef rep_list *list;
int length(const list lst) {
if (lst->head == NULL) {
return 0;
}
else {
lst->head = lst->head->next;
return 1 + length(lst);
}
}
This works, but the head of the list the function accepts as a parameter gets changed. I don't know how to fix that.
I'm not allowed to change the function definition so it should always accept a list variable.
Any ideas?
EDIT: I tried to do what Tyler S suggested in the comments but I encountered another problem. If I create a node* variable at the beginning, it should point to lst->head. But then every recursive call to the function changes the value back to lst->head and I cannot move forward.
You don't need a local node: just don't change the list head. Instead, pass the next pointer as the recursion head.
int length(const list lst) {
if (lst->head == NULL) {
return 0;
}
else {
return 1 + length(lst->head-next);
}
}
I see. Okay; this gets a bit clunky because of the chosen representation. You need a temporary variable to contain the remaining list. This iscludes changing the head.
int length(const list lst) {
if (lst->head == NULL) {
return 0;
}
else {
new_lst = new(list)
new_lst->head = lst->head->next;
var result = 1 + length(new_lst);
free(new_lst)
return result
}
}
At each recursion step, you create a new list object, point it to the 2nd element of the current list, and continue. Does this do the job for you?
Although this solution is clunky and I hate it, its the only way I can see to accomplish what you're asking without modifying the method signature. We create a temporary node * as member data of the class and modify it when we start.
struct rep_list {
struct node *head;
struct node *tail;
}
node *temp = NULL;
bool didSetHead = false;
typedef rep_list *list;
int length(const list lst) {
if ((didSetHead) && (lst->head != temp)) {
temp = lst->head;
didSetHead = false;
}
if (temp == NULL) {
didSetHead = true;
return 0;
}
else {
temp = temp->next;
return 1 + length(temp);
}
}
Please note, I haven't tested this code and you may have to play with a bit, but the idea will work.
I need to write a factorial function that takes in one input and returns the factorial of the given number. if the function is passed in to be 0 or less than 0 then the function should return 0.
I am not actually sure how to write this only using the features of PSScript version 1.0 however I just wrote this, please can someone help me.
JAVA -
public static int factorial (int n) {
if (n<0) {
return 0;
}
return (n<2) ? 1 : n *factorial(n-1);
}
I want to know if there is any I could write this so could use this to write a function in PSScript version 1.0
This is what I have done so far ;
func fac (int n) return int {
if (n<0){
return 0;
}
else
{
return (n<2) ? 1 : n *factorial(n-1);
}
}
Based on the language spec you linked to I would guess the recursive factorial function would look like this in your fictional language:
func fac (int n) returns int {
if (n == 0) {
return 1;
} else {
return n * fac(n - 1);
}
}
Maybe it should check for negative arguments too.