Trees

Build a Binary Tree

Create a fixed seven-node binary tree and render its shape.

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.

node links A node stores one value plus references to its left and right children.

Basic Implementation

basic.sh
Replay: real traced execution (multi-file project)
#!/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
render 4
echo

  1. node ← 1, tree ← 1

    1#!/usr/bin/env bash2declare -A val left right
    values this step1node1tree

  2. node ← 3, tree ← 1, 3

    1#!/usr/bin/env bash2declare -A val left right
    values this step3node1, 3tree

  3. node ← 2, tree ← 2(1,3)

    1#!/usr/bin/env bash2declare -A val left right
    values this step2node2(1,3)tree

  4. node ← 5, tree ← 2(1,3), 5

    1#!/usr/bin/env bash2declare -A val left right
    values this step5node2(1,3), 5tree

  5. node ← 7, tree ← 2(1,3), 5, 7

    1#!/usr/bin/env bash2declare -A val left right
    values this step7node2(1,3), 5, 7tree

  6. node ← 6, tree ← 2(1,3), 6(5,7)

    1#!/usr/bin/env bash2declare -A val left right
    values this step6node2(1,3), 6(5,7)tree

  7. node ← 4, tree ← 4(2(1,3),6(5,7))

    1#!/usr/bin/env bash2declare -A val left right
    values this step4node4(2(1,3),6(5,7))tree

  8. stdout ← 4(2(1,3),6(5,7))

    5local id=$16if [[ "$id" == "0" || -z "$id" ]]; then printf "_"; return; fi7if [[ "${left[$id]}" == "0" && "${right[$id]}" == "0" ]]; then printf "%s" "${val[$id]}"; return; fi
    values this step4(2(1,3),6(5,7))stdout4(2(1,3),6(5,7))tree

Complexity

  • Time: O(n)
  • Space: O(n)

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.