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Programming-Idioms

  • Python
  • Rust

Idiom #128 Breadth-first traversing of a tree

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

A tree with 6 nodes, rooted at node 1
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);
            }
        }
    }
}
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)
#include <deque>
#include <functional>
void bfs(Node& root, std::function<void(Node*)> f) {
  std::deque<Node*> node_queue;
  node_queue.push_back(&root);
  while (!node_queue.empty()) {
    Node* const node = node_queue.front();
    node_queue.pop_front();
    f(node);
    for (Node* const child : node->children) {
      node_queue.push_back(child);
    }
  }
}

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