now bfs using queues

lectures/20_trees_III/README.md

@@ -61,32 +61,43 @@ We can also implement operator++ for the ds_set iterator without using the paren
 
 - Write an algorithm to print the nodes in the tree one tier at a time, that is, in a breadth-first manner.
 
-  ```cpp
-  void breadth_first_traverse(Node* root)
-  {
-    int level=0;
-    std::vector<Node*> current_level;
-    std::vector<Node*> next_level;
-    if(root==NULL){return;}
-    current_level.push_back(root);
-    while(current_level.size()!=0)
-    {
-      std::cout<<"level"<<level<<":";
-      for (unsigned i=0; i<current_level.size();i++)
-       {
-         if(current_level[i]->left != NULL)
-          next_level.push_back(current_level[i]->left);
-         if(current_level[i]->right != NULL)
-          next_level.push_back(current_level[i]->right);
-         std::cout<<" "<<current_level[i]->value;
+```cpp
+// the breadth-first traversal function using std::queue
+void breadth_first_traverse(Node* root) {
+    if (root == NULL) {
+        return;
+    }
+
+    std::queue<Node*> node_queue; // queue to store nodes for BFS traversal
+    node_queue.push(root);  // start by pushing the root node
+
+    int level = 0;
+
+    while (!node_queue.empty()) {
+        int level_size = node_queue.size();  // number of nodes at the current level
+        std::cout << "level " << level << ": ";
+
+        for (int i = 0; i < level_size; i++) {
+            Node* current_node = node_queue.front();  // get the front node
+            node_queue.pop();  // remove the node from the queue
+
+            std::cout << current_node->value << " ";  // print the value of the node
+
+            // push the children of the current node to the queue (if they exist)
+            if (current_node->left != NULL) {
+                node_queue.push(current_node->left);
+            }
+            if (current_node->right != NULL) {
+                node_queue.push(current_node->right);
+            }
         }
-       current_level = next_level;
-       level++;
-       next_level.clear();
-       std::cout<<std.endl;
-      }
-     }
-  ```
+        // after we finish the for loop, the only pointers in the queue, are the pointers pointing to nodes of the next level.
+
+        std::cout << std::endl;
+        level++;
+    }
+}
+```
 
 - What is the best/average/worst-case running time of this algorithm? What is the best/average/worst-case
 memory usage of this algorithm? Give a specific example tree that illustrates each case.