Introduction to Tree Breadth First Search Pattern

Tree BFS (Breadth-First Search) traversal is a key technique used to explore nodes in a binary tree level by level, starting from the root. This pattern is important because it ensures that each node at a given level is processed before moving on to the next level, making it ideal for solving problems that require a level-wise approach. BFS traversal is particularly useful when we need to understand the structure of a tree or solve problems that rely on the tree's hierarchy, such as finding the shortest path or performing level order operations.

The overall approach for solving BFS-related problems involves using a queue data structure. We begin by inserting the root node into the queue, then repeatedly process nodes in the order they are removed from the queue. For each node, we add its child nodes to the queue, ensuring that all nodes at the current level are processed before the next level begins. This makes BFS an efficient way to explore each level of a tree in sequence and solve a wide range of problems that depend on level-wise exploration.

Let's understand the Tree BFS (Breadth-First Search) traversal via below problem.

Problem Statement

Given a root of the binary tree, return the sum of all nodes of the binary tree.

Examples

Example 1:

• Input: root = [1, 2, 3]

• Expected Output: 6
• Justification: The tree has three nodes: 1, 2, and 3. Their sum is 1 + 2 + 3 = 6.

Example 2:

• Input: root = [4, 9, 7, 2, 6]

• Expected Output: 28
• Justification: The tree contains the nodes 4, 9, 7, 2, and 6. Their sum is 4 + 9 + 7 + 2 + 6 = 28.

Example 3:

• Input: root = [10, 5, 3, 7, null, null, 9]

• Expected Output: 34
• Justification: The tree contains the nodes 10, 5, 3, 7, and 9. Their sum is 10 + 5 + 3 + 7 + 9 = 34.

Solution

The algorithm uses BFS (Breadth-First Search) to calculate the sum of all nodes in a binary tree. We initialize a queue starting with the root node. While the queue is not empty, we repeatedly remove the front node, add its value to the sum, and then add its left and right children (if they exist) to the queue. This process continues until all nodes are processed. If the tree is empty (i.e., the root is null), the function returns 0. The overall approach ensures that each node is visited level by level, and its value is added to the cumulative sum.

Step-by-Step Algorithm

Step 1: Check if the tree is empty:

• If the root is null, return 0. This step ensures that we handle edge cases where there are no nodes in the tree.

Step 2: Initialize a queue for BFS traversal:

• A Queue is used to store the nodes at each level of the tree.
• We add the root node to the queue. This step prepares the BFS traversal by starting with the root node.

Step 3: Initialize a variable to store the sum:

• A variable sum is initialized to 0 to keep track of the sum of all node values.

Step 4: Perform BFS traversal:

• While the queue is not empty, we perform the following operations for each node in the queue:

  1. Step 4.1: Remove the front node from the queue:

     • The front node is removed using queue.poll(). This allows us to process the current node in the BFS order.

  2. Step 4.2: Add the node’s value to the sum:

     • The value of the current node is added to the sum. This updates the total sum with the current node's value.

  3. Step 4.3: Add the left child (if it exists) to the queue:

     • If the current node’s left child is not null, add it to the queue. This ensures that we process the left child in subsequent BFS steps.

  4. Step 4.4: Add the right child (if it exists) to the queue:

     • If the current node’s right child is not null, add it to the queue. This ensures that we process the right child in subsequent BFS steps.

Step 5: Return the total sum:

• After processing all nodes, return the value of sum. This value represents the sum of all nodes in the tree.

Algorithm Walkthrough

1. Initial State:

   • Queue: [10]
   • Sum: 0

2. Step 1:

   • Remove 10 from the queue and add its value to the sum.
   • Add left child (5) and right child (3) to the queue.
   • Queue: [5, 3]
   • Sum: 10

3. Step 2:

   • Remove 5 from the queue and add its value to the sum.
   • Add left child (7) to the queue (right child is null).
   • Queue: [3, 7]
   • Sum: 15

4. Step 3:

   • Remove 3 from the queue and add its value to the sum.
   • Add right child (9) to the queue (left child is null).
   • Queue: [7, 9]
   • Sum: 18

5. Step 4:

   • Remove 7 from the queue and add its value to the sum.
   • Node 7 has no children, so nothing is added to the queue.
   • Queue: [9]
   • Sum: 25

6. Step 5:

   • Remove 9 from the queue and add its value to the sum.
   • Node 9 has no children, so nothing is added to the queue.
   • Queue: []
   • Sum: 34

7. Final State:

   • The queue is empty, and the total sum of all nodes is 34.

Code

// class TreeNode {
//     int val;
//     TreeNode left;
//     TreeNode right;

//     // Constructor to create a new node
//     TreeNode(int val) {
//         this.val = val;
//         left = null;
//         right = null;
//     }
// }

public class Solution {

  // Function to calculate the sum of all nodes in a binary tree using BFS
  public static int sumOfNodes(TreeNode root) {
    // If the tree is empty, return 0
    if (root == null) {
      return 0;
    }

    // Initialize a queue to store nodes for BFS traversal
    Queue<TreeNode> queue = new LinkedList<>();
    queue.add(root); // Start with the root node
    int sum = 0; // Variable to store the sum of node values

    // Perform BFS traversal
    while (!queue.isEmpty()) {
      TreeNode currentNode = queue.poll(); // Remove and process the front node from the queue
      sum += currentNode.val; // Add the current node's value to the sum

      // If the left child exists, add it to the queue
      if (currentNode.left != null) {
        queue.add(currentNode.left);
      }

      // If the right child exists, add it to the queue
      if (currentNode.right != null) {
        queue.add(currentNode.right);
      }
    }

    return sum;
  }

  public static void main(String[] args) {
    // Example 1
    TreeNode root1 = new TreeNode(1);
    root1.left = new TreeNode(2);
    root1.right = new TreeNode(3);
    System.out.println("Sum of nodes in Example 1: " + sumOfNodes(root1)); // Output: 6

    // Example 2
    TreeNode root2 = new TreeNode(4);
    root2.left = new TreeNode(9);
    root2.right = new TreeNode(7);
    root2.left.left = new TreeNode(2);
    root2.left.right = new TreeNode(6);
    System.out.println("Sum of nodes in Example 2: " + sumOfNodes(root2)); // Output: 28

    // Example 3
    TreeNode root3 = new TreeNode(10);
    root3.left = new TreeNode(5);
    root3.left.left = new TreeNode(3);
    root3.left.right = new TreeNode(7);
    root3.left.right.right = new TreeNode(9);
    System.out.println("Sum of nodes in Example 3: " + sumOfNodes(root3)); // Output: 34
  }
}

Complexity Analysis

• Time Complexity: The time complexity of the code is , where N is the number of nodes in the binary tree. This is because we visit each node exactly once during the BFS traversal, processing its value and checking its children.

• Space Complexity: The space complexity is in the worst case. This happens because the queue will hold up to N/2 nodes at the last level of the tree in a complete binary tree, and we need space to store the nodes as we traverse the tree. Additionally, space is used for the call stack during the traversal, but the queue dominates the space usage.

Real-World Applications of Tree BFS (Breadth-First Search)

1. Level Order Traversal of Organizational Hierarchies: In companies, organizational hierarchies are often represented as trees. BFS can be used to traverse the hierarchy level by level, identifying all employees at the same level (e.g., all managers, all team leads) to perform operations such as payroll processing or team restructuring.

2. File System Searching: File systems are typically represented as tree structures, where folders are parent nodes and files/sub-folders are child nodes. BFS can be used to explore files and folders at each directory level, which is useful in applications like file explorers to show all items in a folder before drilling down into sub-folders.

3. Binary Heap and Priority Queue Implementation: Binary heaps (a type of binary tree) are used to implement priority queues. BFS is used to traverse the heap level by level when adding or removing elements. This ensures the heap properties are maintained efficiently.

4. Network Packet Broadcasting: In networking, BFS is used to broadcast data packets across a tree structure. For example, BFS ensures that all nodes at the same depth (such as routers or switches) are processed before moving to the next depth level. This allows efficient distribution of data across a network.

5. Serializing and Deserializing Trees: BFS is commonly used to serialize and deserialize binary trees. When saving a tree to a file or transferring it over a network, BFS ensures that all nodes at the same level are processed, which allows the tree structure to be rebuilt efficiently.

Now, let's start solving problems on Tree BFS.

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