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 R 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.R
Replay: real traced execution (multi-file project)
node <- function(value, left = NULL, right = NULL) list(value = value, left = left, right = right)
render <- function(n) {
  if (is.null(n)) return("_")
  if (is.null(n$left) && is.null(n$right)) return(as.character(n$value))
  paste0(n$value, "(", render(n$left), ",", render(n$right), ")")
}
sample_tree <- function() node(4, node(2, node(1), node(3)), node(6, node(5), node(7)))
list_string <- function(values) paste0("[", paste(values, collapse = ", "), "]")
cat(render(sample_tree()), "\n", sep = "")

  1. node ← 1, tree ← 1

    1node <- function(value, left = NULL, right = NULL) list(value = value, left = left, right = right)2render <- function(n) {
    values this step1node1tree

  2. node ← 3, tree ← 1, 3

    1node <- function(value, left = NULL, right = NULL) list(value = value, left = left, right = right)2render <- function(n) {
    values this step3node1, 3tree

  3. node ← 2, tree ← 2(1,3)

    1node <- function(value, left = NULL, right = NULL) list(value = value, left = left, right = right)2render <- function(n) {
    values this step2node2(1,3)tree

  4. node ← 5, tree ← 2(1,3), 5

    1node <- function(value, left = NULL, right = NULL) list(value = value, left = left, right = right)2render <- function(n) {
    values this step5node2(1,3), 5tree

  5. node ← 7, tree ← 2(1,3), 5, 7

    1node <- function(value, left = NULL, right = NULL) list(value = value, left = left, right = right)2render <- function(n) {
    values this step7node2(1,3), 5, 7tree

  6. node ← 6, tree ← 2(1,3), 6(5,7)

    1node <- function(value, left = NULL, right = NULL) list(value = value, left = left, right = right)2render <- function(n) {
    values this step6node2(1,3), 6(5,7)tree

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

    1node <- function(value, left = NULL, right = NULL) list(value = value, left = left, right = right)2render <- function(n) {
    values this step4node4(2(1,3),6(5,7))tree

  8. stdout ← 4(2(1,3),6(5,7))

    8list_string <- function(values) paste0("[", paste(values, collapse = ", "), "]")9cat(render(sample_tree()), "\n", sep = "")
    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

  • node <- function(value, left = NULL, right = NULL) list(...) builds each tree node as an R list with value, left, and right fields.
  • sample_tree() directly nests the calls: node(4, node(2, node(1), node(3)), node(6, node(5), node(7))).
  • In that call, the second argument is the left child and the third argument is the right child. So node(2, node(1), node(3)) forms 2(1,3), and node(6, node(5), node(7)) forms 6(5,7).

Replay steps

left side:  1, 3 -> 2(1,3)
right side: 5, 7 -> 6(5,7)
root:       4 links both sides -> 4(2(1,3),6(5,7))
  • render() returns _ for NULL, as.character(n$value) for a leaf, and paste0(n$value, "(", render(n$left), ",", render(n$right), ")") for an internal node.
  • cat(render(sample_tree()), "\n", sep = "") prints exactly 4(2(1,3),6(5,7)).