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Idiom #128 Breadth-first traversing of a tree

Call a function f on every node of a tree, in breadth-first prefix order

Illustration
func (root *Tree) Bfs(f func(*Tree)) {
	if root == nil {
		return
	}
	queue := []*Tree{root}
	for len(queue) > 0 {
		t := queue[0]
		queue = queue[1:]
		f(t)
		queue = append(queue, t.Children...)
	}
}
def BFS(f, root):
	Q = [root]
	while Q:
		n = Q.pop(0)
		f(n)
		for child in n:
			if not n.discovered:
				n.discovered = True
				Q.append(n)
use std::collections::VecDeque;
struct Tree<V> {
    children: Vec<Tree<V>>,
    value: V
}

impl<V> Tree<V> {
    fn bfs(&self, f: impl Fn(&V)) {
        let mut q = VecDeque::new();
        q.push_back(self);

        while let Some(t) = q.pop_front() {
            (f)(&t.value);
            for child in &t.children {
                q.push_back(child);
            }
        }
    }
}
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Idiom created by

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Related idioms

func (root *Tree) Bfs(f func(*Tree)) {
	if root == nil {
		return
	}
	queue := []*Tree{root}
	for len(queue) > 0 {
		t := queue[0]
		queue = queue[1:]
		f(t)
		queue = append(queue, t.Children...)
	}
}
def BFS(f, root):
	Q = [root]
	while Q:
		n = Q.pop(0)
		f(n)
		for child in n:
			if not n.discovered:
				n.discovered = True
				Q.append(n)
struct Tree<V> {
    children: Vec<Tree<V>>,
    value: V
}

impl<V> Tree<V> {
    fn bfs(&self, f: impl Fn(&V)) {
        let mut q = VecDeque::new();
        q.push_back(self);

        while let Some(t) = q.pop_front() {
            (f)(&t.value);
            for child in &t.children {
                q.push_back(child);
            }
        }
    }
}