medium Find Largest Value in Each Tree Row

Problem Statement

Given the root of a binary tree, return an array containing the largest value in each row of the tree (0-indexed).

Examples

Example 1

• Input: root = [1, 2, 3, 4, 5, null, 6]
• Expected Output: [1, 3, 6]

• Justification:
  • The first row contains 1. The largest value is 1.
  • The second row has 2 and 3, and the largest is 3.
  • The third row has 4, 5, and 6, and the largest is 6.

Example 2

• Input: root = [7, 4, 8, 2, 5, null, 9, null, 3]
• Expected Output: [7, 8, 9, 3]

• Justification:
  • The first row contains 7, and the largest value is 7.
  • The second row has 4 and 8, and the largest is 8.
  • The third row has 2, 5, and 9, and the largest is 9.
  • The fourth row has 3, and the largest is 3.

Example 3

• Input: root = [10, 5]
• Expected Output: [10, 5]
• Justification:
  • The first row has 10, and the largest value is 10.
  • The second row contains 5, and the largest is 5.

Constraints:

• The number of nodes in the tree will be in the range [0, 10⁴].
• -2³¹ <= Node.val <= 2³¹ - 1

Solution

To solve this problem, we need to determine the maximum value at each level (row) of the binary tree. A level order traversal (breadth-first search) is ideal for this task. By traversing each level of the tree one by one, we can find the maximum value in each row and add it to the result list. This approach ensures that every level is fully processed before moving to the next.

Using level order traversal is efficient because it only requires a single pass through the tree, processing each node exactly once. This makes it an optimal solution with a linear time complexity, proportional to the number of nodes in the tree.

Step-by-Step Algorithm

1. Initialize Result List:

   • Create result as an empty list to store the largest values of each row.

2. Check for Empty Tree:

   • If root is null, return the empty result list.

3. Initialize Queue:

   • Create a queue and add root to start level order traversal.

4. Process Each Level:

   • While the queue is not empty:
     • Determine the number of nodes at the current level (levelSize).
     • Set maxVal to Integer.MIN_VALUE.

5. Process Each Node at Current Level:

   • For each node in the current level:
     • Dequeue a node, update maxVal with the maximum of maxVal and the node's value.
     • If the node has a left child, enqueue it.
     • If the node has a right child, enqueue it.

6. Store Largest Value:

   • After processing all nodes at the current level, add maxVal to result.

7. Repeat for All Levels:

   • Continue until the queue is empty.

8. Return Result:

   • Return result containing the largest values in each row.

Algorithm Walkthrough

Input: root = [1, 2, 3, 4, 5, null, 6]

1. Initialize:

   • result = [] (to store largest values)
   • queue = [1] (start with the root node)

2. Level 1:

   • levelSize = 1 (1 node at this level), maxVal = -∞ (initialize to smallest value)
   • Process Node 1:
     • Dequeue node 1 → queue = []
     • maxVal = max(-∞, 1) = 1 (update because -∞ is less than 1)
     • Enqueue children 2 and 3 → queue = [2, 3]
   • Store Largest Value of Level 1:
     • Add 1 to result → result = [1]

3. Level 2:

   • levelSize = 2 (2 nodes at this level), maxVal = -∞ (reset for new level)
   • Process Node 2:
     • Dequeue node 2 → queue = [3]
     • maxVal = max(-∞, 2) = 2 (update because -∞ is less than 2)
     • Enqueue children 4 and 5 → queue = [3, 4, 5]
   • Process Node 3:
     • Dequeue node 3 → queue = [4, 5]
     • maxVal = max(2, 3) = 3 (update because 3 is greater than 2)
     • Enqueue child 6 → queue = [4, 5, 6]
   • Store Largest Value of Level 2:
     • Add 3 to result → result = [1, 3]

4. Level 3:

   • levelSize = 3 (3 nodes at this level), maxVal = -∞ (reset for new level)
   • Process Node 4:
     • Dequeue node 4 → queue = [5, 6]
     • maxVal = max(-∞, 4) = 4 (update because -∞ is less than 4)
   • Process Node 5:
     • Dequeue node 5 → queue = [6]
     • maxVal = max(4, 5) = 5 (update because 5 is greater than 4)
   • Process Node 6:
     • Dequeue node 6 → queue = []
     • maxVal = max(5, 6) = 6 (update because 6 is greater than 5)
   • Store Largest Value of Level 3:
     • Add 6 to result → result = [1, 3, 6]

5. Return Result:

   • Final output: [1, 3, 6] (largest values from each row).

Code

import java.util.*;

// Definition for a binary tree node
// class TreeNode {
//     int val;
//     TreeNode left;
//     TreeNode right;

//     TreeNode(int val) {
//         this.val = val;
//     }
// }

class Solution {

  // Method to find the largest value in each row of the binary tree
  public List<Integer> largestValues(TreeNode root) {
    List<Integer> result = new ArrayList<>();

    // Return an empty list if the root is null
    if (root == null) return result;

    // Initialize a queue for level order traversal
    Queue<TreeNode> queue = new LinkedList<>();
    queue.add(root);

    // Perform level order traversal
    while (!queue.isEmpty()) {
      int levelSize = queue.size();
      int maxVal = Integer.MIN_VALUE;

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

        // Find the maximum value at the current level
        maxVal = Math.max(maxVal, node.val);

        // Add left and right children to the queue for the next level
        if (node.left != null) queue.add(node.left);
        if (node.right != null) queue.add(node.right);
      }

      // Store the largest value of the current level
      result.add(maxVal);
    }

    return result;
  }

  public static void main(String[] args) {
    Solution solution = new Solution();

    // Example 1: Creating tree for [1, 2, 3, 4, 5, null, 6]
    TreeNode root1 = new TreeNode(1);
    root1.left = new TreeNode(2);
    root1.right = new TreeNode(3);
    root1.left.left = new TreeNode(4);
    root1.left.right = new TreeNode(5);
    root1.right.right = new TreeNode(6);

    List<Integer> output1 = solution.largestValues(root1);
    System.out.println("Example 1 Output: " + output1); // Expected Output: [1, 3, 6]

    // Example 2: Creating tree for [7, 4, 8, 2, 5, null, 9, null, 3]
    TreeNode root2 = new TreeNode(7);
    root2.left = new TreeNode(4);
    root2.right = new TreeNode(8);
    root2.left.left = new TreeNode(2);
    root2.left.right = new TreeNode(5);
    root2.right.right = new TreeNode(9);
    root2.left.left.right = new TreeNode(3);

    List<Integer> output2 = solution.largestValues(root2);
    System.out.println("Example 2 Output: " + output2); // Expected Output: [7, 8, 9, 3]

    // Example 3: Creating tree for [10, 5] using the helper method
    TreeNode root3 = new TreeNode(10);
    root3.left = new TreeNode(5);

    List<Integer> output3 = solution.largestValues(root3);
    System.out.println("Example 3 Output: " + output3); // Expected Output: [10, 5]
  }
}

Complexity Analysis

Time Complexity

• The algorithm performs a level-order traversal (BFS) of the binary tree. Each node in the tree is visited exactly once, and all nodes are added to the queue at most once.
• Therefore, the time complexity is , where n is the number of nodes in the binary tree.

Space Complexity

• The space complexity is mainly determined by the queue used for the level-order traversal. In the worst case, the queue could hold all nodes at the deepest level of the tree. For a balanced binary tree, the maximum number of nodes at the deepest level is approximately n/2, where n is the total number of nodes.
• Therefore, the space complexity is in the worst case.