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

Basic Implementation

basic.cpp 
#include <iostream>
#include <queue>
#include <sstream>
#include <string>
#include <vector>
using namespace std;
struct Node { int value; Node* left; Node* right; Node(int v, Node* l=nullptr, Node* r=nullptr): value(v), left(l), right(r) {} };
string render(Node* node) {
    if (node == nullptr) return "_";
    if (node->left == nullptr && node->right == nullptr) return to_string(node->value);
    return to_string(node->value) + "(" + render(node->left) + "," + render(node->right) + ")";
}
Node* sampleTree() {
    return new Node(4, new Node(2, new Node(1), new Node(3)), new Node(6, new Node(5), new Node(7)));
}
string listString(const vector<int>& values) {
    stringstream out; out << "[";
    for (size_t i = 0; i < values.size(); i++) { if (i) out << ", "; out << values[i]; }
    out << "]"; return out.str();
}
bool search(Node* root, int target) { Node* node = root; while (node) { if (target == node->value) return true; node = target < node->value ? node->left : node->right; } return false; }
int main() { Node* root = sampleTree(); cout << (search(root, 5) ? "5 found" : "5 not found") << "\n"; cout << (search(root, 8) ? "8 found" : "8 not found") << "\n"; }

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.