LeetCode 2583: Kth Largest Sum in a Binary Tree – Solution Explained

When working with binary trees, one common task is to analyze the properties of the tree's levels. In this blog post, we'll walk through a solution to LeetCode Problem 2583: Kth Largest Sum in a Binary Tree, a problem that challenges us to compute and find specific sums from different levels of a binary tree.

Problem Statement

You are given the root of a binary tree and a positive integer k. Your goal is to return the k-th largest level sum in the tree.

The level sum is defined as the sum of the values of all nodes at the same depth in the tree. If the number of levels in the tree is less than k, the function should return -1.

Example 1:

• Input: root = [5, 8, 9, 2, 1, 3, 7, 4, 6], k = 2
• Output: 13
• Explanation:
  • Level 1 sum: 5
  • Level 2 sum: 8 + 9 = 17
  • Level 3 sum: 2 + 1 + 3 + 7 = 13
  • Level 4 sum: 4 + 6 = 10
  • The 2nd largest sum is 13.

Example 2:

• Input: root = [1, 2, null, 3], k = 1
• Output: 3
• Explanation:
  • The largest level sum is 3 (at the third level).

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This problem essentially boils down to two tasks:

1. Traverse the binary tree to compute the sum of each level.
2. Find the k-th largest sum from the computed sums.

If you're unfamiliar with binary trees, each "level" refers to the nodes at the same depth. For example, the root is at level 1, the root's children are at level 2, and so on.

Approach to Solve the Problem

We can break down the solution into three key steps:

1. Perform a Level-Order Traversal (BFS): This will help us compute the sum of values at each level of the tree.
2. Store Level Sums in a Min-Heap (Priority Queue): The use of a min-heap allows us to efficiently track the k largest sums.
3. Return the k-th Largest Level Sum: Once the traversal is done, we extract the k-th largest sum from the heap.

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Efficient Solution Explanation

We will use a Breadth-First Search (BFS) to traverse the tree level by level, computing the sum of node values at each level. To keep track of the largest sums, we will use a min-heap (priority queue) of size k. This way, we can maintain only the top k sums and efficiently return the k-th largest one.

Here is the optimized solution: (cpp)

#include <queue>
#include <vector>
using namespace std;

class Solution
{
public:
    long long kthLargestLevelSum(TreeNode *root, int k)
    {
        // Queue for BFS traversal
        queue<TreeNode *> q;
        q.push(root);

        // Min-heap to store the top k largest sums
        priority_queue<long long, vector<long long>, greater<long long>> pq;

        while (!q.empty())
        {
            int size = q.size();
            long long levelSum = 0;

            // Compute sum for the current level
            for (int i = 0; i < size; ++i)
            {
                TreeNode *node = q.front();
                q.pop();
                levelSum += node->val;

                if (node->left)
                    q.push(node->left);
                if (node->right)
                    q.push(node->right);
            }

            // Insert level sum into the heap
            if (pq.size() < k)
            {
                pq.push(levelSum);
            }
            else if (levelSum > pq.top())
            {
                pq.pop();
                pq.push(levelSum);
            }
        }

        // If there are fewer than k levels, return -1
        if (pq.size() < k)
            return -1;

        return pq.top(); // The k-th largest sum
    }
};

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Step-by-Step Breakdown

1. Level-Order Traversal (BFS):

   • We use a queue to perform a level-order traversal (BFS). This allows us to process one level at a time and compute the sum of node values at each level.

2. Min-Heap for Top k Sums:

   • A min-heap of size k is used to track the top k largest level sums.
   • If the heap has fewer than k elements, we simply add the sum of the current level.
   • If the heap already has k elements and the current level sum is larger than the smallest sum in the heap, we replace the smallest sum with the current level sum.

3. Return the k-th Largest Sum:

   • After processing all the levels, the smallest element in the heap will be the k-th largest level sum.

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Complexity Analysis

• Time Complexity:

  • Traversing all nodes takes O(n), where n is the number of nodes in the tree.
  • Inserting into and maintaining the heap takes O(log k) for each level.
  • The overall time complexity is O(n log k).

• Space Complexity:

• The queue for BFS requires O(n) space in the worst case.
• The heap requires O(k) space.
• Thus, the total space complexity is O(n + k).

This problem is a great example of how data structures like heaps can optimize solutions involving ranked selections (like finding the k-th largest sum). By combining BFS with a min-heap, we can efficiently solve the problem without unnecessary memory overhead.

Whether you're practicing for coding interviews or just improving your tree traversal skills, this problem is a fun challenge to sharpen your binary tree knowledge!

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Key Takeaways:

1. Binary Tree Traversal: Understand level-order traversal (BFS) to process levels in a tree.
2. Min-Heap (Priority Queue): Learn how to use heaps to track the top k largest elements efficiently.
3. Time and Space Complexity: Aim for solutions that balance time and space efficiency for large inputs.

Feel free to implement this solution in your practice and explore how this optimization technique can be applied to other ranked problems!

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FAQ:

• Q: Can I use a different data structure instead of a heap?

  • A: Yes, but using a heap ensures that we maintain the top k sums efficiently. Using an array or other data structures would either increase time complexity or require additional sorting steps.

• Q: What if there are fewer than k levels in the tree?

  • A: In that case, the solution returns -1 as instructed in the problem.

By understanding how to approach problems involving binary trees and level sums, you're one step closer to mastering tree-based coding challenges!

This blog post covers the problem 2583. Kth Largest Sum in a Binary Tree in a simple and clear manner, helping users understand the solution step-by-step.