Visit a tree breadth-first with a queue.

Algorithm

The canonical tree is 4(2(1,3),6(5,7)), so this C# DSA implementation can be compared directly with the rest of the DSA track.

Basic Implementation

basic.cs
using System;
using System.Collections.Generic;
using System.Linq;

class Node {
    public int Value;
    public Node? Left;
    public Node? Right;
    public Node(int value, Node? left = null, Node? right = null) { Value = value; Left = left; Right = right; }
}
class Program {
    static string Render(Node? node) {
        if (node == null) return "_";
        if (node.Left == null && node.Right == null) return node.Value.ToString();
        return $"{node.Value}({Render(node.Left)},{Render(node.Right)})";
    }
    static Node SampleTree() => new Node(4, new Node(2, new Node(1), new Node(3)), new Node(6, new Node(5), new Node(7)));
    static string ListString(IEnumerable<int> values) => "[" + string.Join(", ", values) + "]";
    static void Main() { var queue = new Queue<Node>(); queue.Enqueue(SampleTree()); var output = new List<int>(); while (queue.Count > 0) { var node = queue.Dequeue(); output.Add(node.Value); if (node.Left != null) queue.Enqueue(node.Left); if (node.Right != null) queue.Enqueue(node.Right); } Console.WriteLine(ListString(output)); }
}

Complexity

  • Time: O(n)
  • Space: O(w) queue space

Implementation notes

  • Render tree structure explicitly instead of printing node objects.
  • The replay highlights the node, traversal state, queue, path, or search cursor that changes at each step.
level order Level-order traversal uses a queue to visit shallower nodes first.