Introduction to Tree View Pattern

Tree view problems focus on understanding how a binary tree appears from different perspectives, such as the left, right, and top views. These problems are important because they help in visualizing the tree structure in ways that can be applied to real-world problems like printing the skyline of buildings or managing layered data structures. By learning this pattern, programmers gain a deeper understanding of how trees can be processed based on their visual representation.

To solve Tree view problems, we usually perform level-order traversal to explore the nodes at each level, but we also track nodes based on their position in the tree. For example, in the left view, we only consider the first node visible from the left side at each level. Similarly, in the right view, we look at the rightmost node. The overall approach relies on understanding the position of nodes and efficiently traversing them to capture the correct view.

Now, let's look at the example below to find the left view of the given binary tree.

Problem Statement

Given a root of the binary tree, print the left view of the tree.

The left view of a binary tree consists of nodes that are visible when the tree is viewed from the left side. For each level of the tree, the first node encountered in that level will be included in the left view.

Examples

Example 1

• Input: root = [1, 2, 3, 4, 5]

• Expected Output: [1, 2, 4]
• Justification:
  • The first node at level 0 is 1.
  • The first node at level 1 is 2.
  • The first node at level 2 is 4.

Example 2

• Input: root = [10, 7, 12, 6, null, null, null, 4]

• Expected Output: [10, 7, 6, 4]
• Justification:
  • The first node at level 0 is 10.
  • The first node at level 1 is 7.
  • The first node at level 2 is 6.
  • The first node at level 3 is 4.

Example 3

• Input: root = [8, 4, 9, 3, null, null, 10, 2]

• Expected Output: [8, 4, 3, 2]
• Justification:
  • The first node at level 0 is 8.
  • The first node at level 1 is 4.
  • The first node at level 2 is 3.
  • The first node at level 3 is 2.

Solution

The algorithm uses Breadth-First Search (BFS) to traverse the tree level by level and identify the leftmost node at each level. It starts by adding the root node to a queue and processes each level of the tree iteratively. For each level, the first node encountered is the leftmost node and is included in the result. At each step, the left child of the node is added to the queue first, followed by the right child, ensuring that the leftmost node is encountered first at every level.

Step-by-Step Algorithm

1. Base Case: If the tree is empty (root == null), return immediately since there is no left view to compute.
2. Initialize Queue: Create an empty queue to help with the level-order traversal. Add the root node to the queue. This is done because we need to process the tree level by level, starting from the root.
3. While Queue is Not Empty:
   • Level Processing: Determine the number of nodes at the current level using queue.size(). This ensures that we only process nodes that belong to the same level.
   • Traverse Level:
     • For each node at the current level:
       • Check First Node: If it is the first node of the level (i == 0), it is part of the left view, so print or store its value.
       • Add Left Child: If the node has a left child, add it to the queue. This is done to ensure the leftmost node is processed first at the next level.
       • Add Right Child: If the node has a right child, add it to the queue. This ensures all nodes of the current level are processed before moving to the next level.
4. Repeat: Continue processing nodes in the queue until all levels of the tree are processed and all leftmost nodes are identified.

Algorithm Walkthrough

Input: root = [10, 7, 12, 6, null, null, null, 4]:

Tree Structure:

        10
       /  \
      7    12
     /  
    6  
   /
  4

1. Initialization: The root node is 10. We add 10 to the queue.

   Queue: [10]

2. Level 1:

   • queue.size() is 1, meaning there is 1 node at this level.
   • Dequeue node 10. Since it's the first node of the level, it is part of the left view, so print 10.
   • Add its left child (7) and right child (12) to the queue.

   Output: 10 Queue: [7, 12]

3. Level 2:

   • queue.size() is 2, meaning there are 2 nodes at this level.
   • Dequeue node 7. Since it's the first node of the level, it is part of the left view, so print 7.
   • Add its left child (6) to the queue (it has no right child).
   • Dequeue node 12. It’s not the first node at this level, so we don't print it.

   Output: 10, 7 Queue: [6]

4. Level 3:

   • queue.size() is 1, meaning there is 1 node at this level.
   • Dequeue node 6. Since it's the first node of the level, it is part of the left view, so print 6.
   • Add its left child (4) to the queue (it has no right child).

   Output: 10, 7, 6 Queue: [4]

5. Level 4:

   • queue.size() is 1, meaning there is 1 node at this level.
   • Dequeue node 4. Since it's the first node of the level, it is part of the left view, so print 4.
   • Node 4 has no children, so nothing is added to the queue.

   Output: 10, 7, 6, 4 Queue: []

6. Completion: The queue is now empty, meaning all levels of the tree have been processed, and the left view of the tree is complete.

Final Output: [10, 7, 6, 4]

Code

import java.util.LinkedList;
import java.util.Queue;

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

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

public class Solution {

  public void leftView(TreeNode root) {
    // Base case: If the tree is empty
    if (root == null) return;

    // Use a queue for BFS
    Queue<TreeNode> queue = new LinkedList<>();
    queue.add(root);

    // Traverse the tree level by level
    while (!queue.isEmpty()) {
      int levelSize = queue.size(); // Number of nodes at the current level

      // Traverse all nodes at the current level
      for (int i = 0; i < levelSize; i++) {
        TreeNode currentNode = queue.poll();

        // The first node of the level is the left view node
        if (i == 0) {
          System.out.print(currentNode.val + " ");
        }

        // Add left child to the queue
        if (currentNode.left != null) {
          queue.add(currentNode.left);
        }

        // Add right child to the queue
        if (currentNode.right != null) {
          queue.add(currentNode.right);
        }
      }
    }
  }

  public static void main(String[] args) {
    // Example 1: Tree = [1, 2, 3, 4, 5]
    TreeNode root = new TreeNode(1);
    root.left = new TreeNode(2);
    root.right = new TreeNode(3);
    root.left.left = new TreeNode(4);
    root.left.right = new TreeNode(5);

    // Print the left view of the tree
    Solution sol = new Solution();
    System.out.print("Left view: ");
    sol.leftView(root); // Output: 1 2 4
  }
}

Complexity Analysis

Time Complexity

The algorithm visits each node exactly once, making the time complexity , where N is the total number of nodes in the tree. Every node is processed in constant time during traversal.

Space Complexity

The space complexity depends on the number of nodes stored in the queue. In the worst case (balanced tree), the queue holds nodes, while in the best case (skewed tree), it holds nodes, where H is the height of the tree.

Now, let's start solving problems on Tree View pattern.

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