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

Basic Implementation

basic.f90

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)
  front = 1; back = 1; queue(1) = 4; n = 0
  do while (front <= back)
    id = queue(front); front = front + 1; n = n + 1; output(n) = value(id)
    if (left(id) /= 0) then; back = back + 1; queue(back) = left(id); end if
    if (right(id) /= 0) then; back = back + 1; queue(back) = right(id); end if
  end do
  call print_list(output, n)
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

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.