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

Basic Implementation

basic.dart 
class Node {
  Node(this.value, [this.left, this.right]);
  final int value;
  Node? left;
  Node? right;
}
String render(Node? node) {
  if (node == null) return "_";
  if (node.left == null && node.right == null) return node.value.toString();
  return "${node.value}(${render(node.left)},${render(node.right)})";
}
Node sampleTree() => Node(4, Node(2, Node(1), Node(3)), Node(6, Node(5), Node(7)));
String listString(List<int> values) => "[${values.join(", ")}]";
bool search(Node? root, int target) { var node = root; while (node != null) { if (target == node.value) return true; node = target < node.value ? node.left : node.right; } return false; }
void main() { final root = sampleTree(); print(search(root, 5) ? '5 found' : '5 not found'); print(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.