Trees

BST Insert

Insert values into a binary search tree by comparing at each node.

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.

binary search tree Values smaller than a node go left; larger values go right.

Visual walkthrough

BST insertion is a comparison path. The pinned tree 4(2(1,3),6(5,7)) is shown with the inserted value taking its sorted slot.

Step 1 - Start at root

For value 5, compare with 4 first; 5 is larger, so move right.

First comparison: 5 > 4, so the search for the insert slot goes right.insert 54compare26137

Step 2 - Take the left slot under 6

At 6, value 5 is smaller, so it becomes the left child.

Second comparison: 5 < 6, so the open left slot is used.426compare135new7

Step 3 - Canonical tree

The resulting tree is the pinned shape 4(2(1,3),6(5,7)).

Final BST after 5 is present under 6.4261357

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)
  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

Complexity

  • Time: O(h) per insert
  • Space: O(n)

Implementation notes

  • Render tree structure explicitly instead of printing node objects.
  • The executable builds the canonical balanced tree. The replay also includes a sorted-order contrast where height grows to 4, showing why an unbalanced BST can degrade to O(n) without rotation.