Problem Description

This LeetCode problem requires finding the k^th largest level sum in a binary tree. A level sum is defined as the total value of all nodes at the same depth, where depth is measured by the distance from the root node. We are given the root of the binary tree and a positive integer k, and we need to return the k^th largest value among all level sums.

If the tree doesn't have k levels, we must return -1. This problem emphasizes that the notion of level is based on a node's distance from the root, meaning that all nodes at the same distance (or depth) from the root are on the same level.

Flowchart Walkthrough

Let's analyze the problem using the Flowchart step by step to determine the appropriate algorithm for LeetCode 2583. Kth Largest Sum in a Binary Tree:

Is it a graph?

• Yes: A binary tree is a special case of a graph where each node has at most two children.

Is it a tree?

• Yes: A binary tree is by definition a tree.

Does the problem relate to directed acyclic graphs (DAGs)?

• No: We are dealing with a binary tree structure, not a general directed acyclic graph.

Does the problem involve shortest paths?

• No: The task is to find the kth largest sum path, not the shortest path between nodes.

Does the problem involve connectivity?

• No: The problem isn't about finding connected components but rather paths and their sums.

Is the graph weighted?

• Not necessarily stated, but in this context, we're not dealing with graph weights in the conventional sense of shortest path problems. The values of nodes can be considered more as values along a path rather than weights affecting traversal.

Conclusion: Despite not directly leading to BFS, understanding and exploring tree structures often involves BFS for level-order traversal or to explore all possible paths. While the flowchart doesn't point directly to BFS for finding sums of paths, BFS could potentially be used to traverse the tree, visiting all nodes and calculating path sums, especially if combined with a priority queue to track kth largest sums. However, based on the flowchart, the path for BFS isn't explicitly outlined for this specific type of problem without further adaptation or additional context.

Intuition

To solve this problem, we will perform a level-order traversal (also known as breadth-first search) to calculate the sum of each level in the tree. Since we traverse the tree level by level, we can easily compute the sum of the nodes at each depth. The level-order traversal uses a queue to keep track of the nodes as we progress.

Here's the step-by-step approach to arriving at the solution:

1. Initialize an empty list arr to store the sum of the values at each level and a queue q initialized with the root node of the tree.
2. While the queue is not empty, we proceed to the following:
   • Initialize a temporary sum t to 0 to accumulate the sum of the values for the current level.
   • For all the nodes in the queue (which are all on the same level):
     • Dequeue the node from the front of the queue.
     • Add the node's value to t.
     • If the node has a left child, enqueue it to the queue.
     • If the node has a right child, enqueue it to the queue.
   • After processing all the nodes at the current level, append the sum t to the list arr.
3. After the traversal is complete, we use the nlargest function from Python's heapq module to fetch the k highest values from the array. The last element in the list returned by nlargest will be the k^th largest value.
4. Finally, we check if the array contains fewer than k elements. If so, return -1 as there are not enough levels in the tree. Otherwise, we return the last element from the list obtained in the previous step, which is the k^th largest level sum.

Following this intuitive approach provides us with a simple and effective solution to identify the k^th largest level sum in the given binary tree.

Learn more about Tree, Breadth-First Search and Binary Search patterns.

Solution Approach

The solution follows a breadth-first search (BFS) approach, which is perfectly suited for level-by-level traversal in a binary tree. The key data structures and algorithms used are:

• A deque from the collections module: It functions as a double-ended queue allowing for efficient insertion and removal from both ends. In our BFS, it's used to enqueue and dequeue nodes.
• A list arr to keep the sum of values for each level.
• The nlargest function from Python's heapq module: Heaps are used to obtain the largest (or smallest) elements without sorting the entire list, and nlargest is specifically optimized to find the 'n' largest elements in a dataset.

Here's a detailed breakdown of the implementation:

1. Initialize a list arr which will hold the sums of node values at each level and a deque called q with the root as the only element. The deque structure allows easy access for enqueueing and dequeueing nodes as we traverse the tree level by level.

2. Initiate a while loop that continues as long as there are nodes in the queue q. This loop continues until every node has been visited.

3. Inside the loop, set a temporary cumulative sum t to zero before considering each level. This is crucial for accumulating the sum of node values at the current level.

4. Execute another loop over q for the number of nodes at the current level (determined by len(q)). Here are the key steps within this loop:

   • Dequeue (popleft()) the current node from q and add its value to the temporary sum t.
   • Check if the current node has a left child (root.left). If so, enqueue this child node to q.
   • Similarly, check for a right child (root.right) and enqueue it if present.

5. After processing all nodes at the current level, append the level sum t to the list arr.

6. Once the while loop is complete (no more nodes to process in the queue), check if we have at least k levels by comparing the length of arr with k.

7. If there aren't enough levels, we return -1 as specified. Otherwise, we find the k^th largest value in the sums collected by calling nlargest(k, arr)[-1]. The nlargest function constructs a min-heap of the k largest elements from arr and returns them as a list. We are only interested in the k^th element, which is the last one in the list returned by nlargest.

This implementation ensures efficient computation of the solution, leveraging BFS for traversal and a min-heap to effectively get the k^th largest value without the need to sort the whole arr, which could be more expensive in terms of time complexity.

Example Walkthrough

Let's illustrate the solution approach using a simple example:

Consider the following binary tree:

         5
       /   \
      3     8
     / \   / \
    2   4 6   9
  
And let k = 2, meaning we want to find the 2nd largest level sum.

1. Initialize a list arr = []. Initialize a queue q and put the root node (value 5) in it.

2. Begin the level-order traversal:

   • Level 0 sum: Queue = [5]

     • Dequeue 5, add to temporary sum t = 5. Enqueue its left (3) and right (8) children.
     • Level sum = 5, so arr = [5].

   • Level 1 sum: Queue = [3, 8]

     • Dequeue 3, add to t = 3. Enqueue its left (2) and right (4) children.
     • Dequeue 8, add to t = 11. Enqueue its left (6) and right (9) children.
     • Level sum = 11, arr = [5, 11].

   • Level 2 sum: Queue = [2, 4, 6, 9]

     • Dequeue 2, add to t = 2.
     • Dequeue 4, add to t = 6.
     • Dequeue 6, add to t = 12.
     • Dequeue 9, add to t = 21.
     • Level sum = 21, arr = [5, 11, 21].

3. After the traversal, arr = [5, 11, 21] now contains the level sums.

4. We use nlargest to fetch the 2 largest values from arr: nlargest(2, arr) gives [21, 11].

5. Since arr contains more than k elements, we do not return -1.

6. Finally, we return the last element in the list returned by nlargest which is the 2nd largest value, 11.

By following these steps, we efficiently solve the problem and find the 2nd largest level sum in our example binary tree.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(N + klogN), where N is the number of nodes in the binary tree. The O(N) term comes from the BFS traversal of the tree, which processes each node once. The O(klogN) term comes from finding the k-th largest sum using the nlargest function from the heap module, which is typically implemented as a heap-based selection algorithm with that complexity.

The space complexity is O(N), as the queue q can contain at most all the nodes of the last level in the worst case (considering a complete binary tree), and the array arr will also contain a sum for each level of the tree, which can be at most N if the tree is skewed (i.e., each level contains only one node).

Learn more about how to find time and space complexity quickly using problem constraints.