BST Search - Dart DSA

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.

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

Visual walkthrough

A BST search follows one comparison path. The same pinned tree shows a found path for 5 and a missing path for 8.

Step 1 - Find 5

Search 5 takes right from 4, then left from 6, then matches 5.

Step 2 - Miss 8

Search 8 takes right from 4, right from 6, right from 7, then reaches null.

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.