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 Bash DSA implementation can be compared directly with the rest of the DSA track.

Basic Implementation

basic.sh

#!/usr/bin/env bash
declare -A val left right
new_node() { local id=$1 value=$2 l=${3:-0} r=${4:-0}; val[$id]=$value; left[$id]=$l; right[$id]=$r; }
render() {
  local id=$1
  if [[ "$id" == "0" || -z "$id" ]]; then printf "_"; return; fi
  if [[ "${left[$id]}" == "0" && "${right[$id]}" == "0" ]]; then printf "%s" "${val[$id]}"; return; fi
  printf "%s(" "${val[$id]}"; render "${left[$id]}"; printf ","; render "${right[$id]}"; printf ")"
}
sample_tree() {
  new_node 1 1; new_node 3 3; new_node 2 2 1 3
  new_node 5 5; new_node 7 7; new_node 6 6 5 7
  new_node 4 4 2 6
}
list_string() {
  local joined=""
  for value in "$@"; do
    [[ -n "$joined" ]] && joined+=", "
    joined+="$value"
  done
  printf '[%s]' "$joined"
}
sample_tree
queue=(4); output=(); front=0
while (( front < ${#queue[@]} )); do id=${queue[$front]}; ((front++)); output+=("${val[$id]}"); [[ "${left[$id]}" != "0" ]] && queue+=("${left[$id]}"); [[ "${right[$id]}" != "0" ]] && queue+=("${right[$id]}"); done
list_string "${output[@]}"; echo

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.