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

Basic Implementation

basic.ts 
class Node {
  value: number;
  left: Node | null;
  right: Node | null;
  constructor(value: number, left: Node | null = null, right: Node | null = null) {
    this.value = value;
    this.left = left;
    this.right = right;
  }
}
function render(node: Node | null): string {
  if (node === null) return "_";
  if (node.left === null && node.right === null) return String(node.value);
  return `${node.value}(${render(node.left)},${render(node.right)})`;
}
function sampleTree(): Node {
  return new Node(4, new Node(2, new Node(1), new Node(3)), new Node(6, new Node(5), new Node(7)));
}

const root = sampleTree();
function search(root, target) { let node = root; while (node !== null) { if (target === node.value) return true; node = target < node.value ? node.left : node.right; } return false; }
console.log(search(root, 5) ? "5 found" : "5 not found");
console.log(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.