Level-Order Traversal

Visit a tree breadth-first with a queue.

Algorithm

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

Basic Implementation

basic.go

package main

import (
    "fmt"
    "strings"
)

type Node struct { value int; left *Node; right *Node }
func render(node *Node) string {
    if node == nil { return "_" }
    if node.left == nil && node.right == nil { return fmt.Sprintf("%d", node.value) }
    return fmt.Sprintf("%d(%s,%s)", node.value, render(node.left), render(node.right))
}
func sampleTree() *Node {
    return &Node{4, &Node{2, &Node{1, nil, nil}, &Node{3, nil, nil}}, &Node{6, &Node{5, nil, nil}, &Node{7, nil, nil}}}
}
func listString(values []int) string {
    parts := []string{}
    for _, value := range values { parts = append(parts, fmt.Sprintf("%d", value)) }
    return "[" + strings.Join(parts, ", ") + "]"
}
func main() { queue := []*Node{sampleTree()}; output := []int{}; for len(queue) > 0 { node := queue[0]; queue = queue[1:]; output = append(output, node.value); if node.left != nil { queue = append(queue, node.left) }; if node.right != nil { queue = append(queue, node.right) } }; fmt.Println(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.