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 Fortran 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.f90
Replay: real traced execution (multi-file project)
program main
  implicit none
  integer :: value(7), left(7), right(7)
  integer :: queue(8), front, back, output(7), n, id
  character(len=128) :: out
  call setup(value, left, right)
  out = ''
  call render_node(4, value, left, right, out)
  print '(A)', trim(out)
contains
  subroutine setup(value, left, right)
    integer, intent(out) :: value(7), left(7), right(7)
    value = [1,2,3,4,5,6,7]; left = 0; right = 0
    left(4) = 2; right(4) = 6; left(2) = 1; right(2) = 3; left(6) = 5; right(6) = 7
  end subroutine setup
  subroutine append(out, text)
    character(len=*), intent(inout) :: out
    character(len=*), intent(in) :: text
    out = trim(out) // text
  end subroutine append
  recursive subroutine render_node(id, value, left, right, out)
    integer, intent(in) :: id, value(7), left(7), right(7)
    character(len=*), intent(inout) :: out
    character(len=16) :: buf
    if (id == 0) then; call append(out, '_'); return; end if
    write(buf, '(I0)') value(id); call append(out, trim(buf))
    if (left(id) /= 0 .or. right(id) /= 0) then
      call append(out, '('); call render_node(left(id), value, left, right, out); call append(out, ',')
      call render_node(right(id), value, left, right, out); call append(out, ')')
    end if
  end subroutine render_node
  recursive subroutine preorder(id, value, left, right, output, n)
    integer, intent(in) :: id, value(7), left(7), right(7)
    integer, intent(inout) :: output(7), n
    if (id == 0) return
    n = n + 1; output(n) = value(id)
    call preorder(left(id), value, left, right, output, n)
    call preorder(right(id), value, left, right, output, n)
  end subroutine preorder
  subroutine print_list(output, n)
    integer, intent(in) :: output(7), n
    integer :: i
    write(*, '(A)', advance='no') '['
    do i = 1, n
      if (i > 1) write(*, '(A)', advance='no') ', '
      write(*, '(I0)', advance='no') output(i)
    end do
    print '(A)', ']'
  end subroutine print_list
  logical function search_value(id, target, value, left, right)
    integer, intent(in) :: id, target, value(7), left(7), right(7)
    integer :: cur
    cur = id
    do while (cur /= 0)
      if (target == value(cur)) then; search_value = .true.; return; end if
      if (target < value(cur)) then; cur = left(cur); else; cur = right(cur); end if
    end do
    search_value = .false.
  end function search_value
end program main

  1. node ← 1, tree ← 1

    1program main2  implicit none
    values this step1node1tree

  2. node ← 3, tree ← 1, 3

    1program main2  implicit none
    values this step3node1, 3tree

  3. node ← 2, tree ← 2(1,3)

    1program main2  implicit none
    values this step2node2(1,3)tree

  4. node ← 5, tree ← 2(1,3), 5

    1program main2  implicit none
    values this step5node2(1,3), 5tree

  5. node ← 7, tree ← 2(1,3), 5, 7

    1program main2  implicit none
    values this step7node2(1,3), 5, 7tree

  6. node ← 6, tree ← 2(1,3), 6(5,7)

    1program main2  implicit none
    values this step6node2(1,3), 6(5,7)tree

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

    1program main2  implicit none
    values this step4node4(2(1,3),6(5,7))tree

  8. stdout ← 4(2(1,3),6(5,7))

    8  call render_node(4, value, left, right, out)9  print '(A)', trim(out)10contains
    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.