Search a binary search tree for one present and one absent value.

Algorithm

The canonical tree is 4(2(1,3),6(5,7)), so this Rust DSA implementation can be compared directly with the rest of the DSA track.

Basic Implementation

basic.rs
use std::collections::VecDeque;
struct Node { value: i32, left: Option<Box<Node>>, right: Option<Box<Node>> }
impl Node {
    fn new(value: i32) -> Self { Self { value, left: None, right: None } }
    fn with(value: i32, left: Node, right: Node) -> Self {
        Self { value, left: Some(Box::new(left)), right: Some(Box::new(right)) }
    }
}
fn render(node: &Option<Box<Node>>) -> String {
    match node {
        None => "_".to_string(),
        Some(n) => {
            if n.left.is_none() && n.right.is_none() { n.value.to_string() }
            else { format!("{}({},{})", n.value, render(&n.left), render(&n.right)) }
        }
    }
}
fn sample_tree() -> Option<Box<Node>> {
    Some(Box::new(Node::with(4, Node::with(2, Node::new(1), Node::new(3)), Node::with(6, Node::new(5), Node::new(7)))))
}
fn list_string(values: &[i32]) -> String {
    format!("[{}]", values.iter().map(|v| v.to_string()).collect::<Vec<_>>().join(", "))
}
fn search(root: &Option<Box<Node>>, target: i32) -> bool { let mut node = root.as_ref(); while let Some(n) = node { if target == n.value { return true; } node = if target < n.value { n.left.as_ref() } else { n.right.as_ref() }; } false }
fn main() { let root = sample_tree(); println!("{}", if search(&root, 5) { "5 found" } else { "5 not found" }); println!("{}", if search(&root, 8) { "8 found" } else { "8 not found" }); }

Complexity

  • Time: O(h) per search
  • Space: O(1) iterative

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.
search path A comparison chooses one subtree at each step, so whole branches are skipped.