Trees

BST Search

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

Basic Implementation

Basic.java 
import java.util.*;

public class Basic {
    static class Node {
        int value;
        Node left;
        Node right;
        Node(int value) { this.value = value; }
        Node(int value, Node left, Node right) { this.value = value; this.left = left; this.right = right; }
    }
    static String render(Node node) {
        if (node == null) return "_";
        if (node.left == null && node.right == null) return Integer.toString(node.value);
        return node.value + "(" + render(node.left) + "," + render(node.right) + ")";
    }
    static Node sampleTree() {
        return new Node(4, new Node(2, new Node(1), new Node(3)), new Node(6, new Node(5), new Node(7)));
    }
    static boolean search(Node root, int target) { Node node = root; while (node != null) { if (target == node.value) return true; node = target < node.value ? node.left : node.right; } return false; }
    public static void main(String[] args) { Node root = sampleTree(); System.out.println(search(root, 5) ? "5 found" : "5 not found"); System.out.println(search(root, 8) ? "8 found" : "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.