Find the Level of Tree with Minimum Sum

Find the Level of Tree with Minimum Sum

What is this problem about?

In the Find the Level of Tree with Minimum Sum coding problem, you are given a binary tree. You need to calculate the sum of all node values at each level and identify which level has the smallest total sum. If multiple levels have the same minimum sum, you usually return the one with the smallest index (closest to the root). Levels are 1-indexed.

Why is this asked in interviews?

Microsoft asks this question to evaluate a candidate's proficiency with Breadth-First Search (BFS) or Level Order Traversal. It tests your ability to maintain state (sums) while moving through a tree structure and your attention to detail regarding 1-indexing and tie-breaking rules. It’s a standard application of the Binary Tree interview pattern.

Algorithmic pattern used

This problem is best solved using BFS (Breadth-First Search).

1. Initialize a queue with the root node.
2. While the queue is not empty:
   • Get the size of the queue (this is the number of nodes at the current level).
   • Iterate through that many nodes, adding their values to a currentSum.
   • Add their children to the queue for the next level.
3. Compare currentSum with the global minSum and update the minLevel if a new minimum is found.

Example explanation

Tree: Level 1: [10] (Sum 10) Level 2: [2, 3] (Sum 5) Level 3: [8, -1] (Sum 7)

1. Level 1 sum = 10. minSum = 10, minLevel = 1.
2. Level 2 sum = 5. 5 < 10, so minSum = 5, minLevel = 2.
3. Level 3 sum = 7. 7 > 5, no change. Result: 2.

Common mistakes candidates make

• Negative Values: Forgetting that sums can be negative, so initializing minSum to 0 might lead to incorrect results. Use Infinity or the first level's sum.
• DFS depth tracking: Trying to use DFS and forgetting to handle level indices correctly in an array or map.
• Off-by-one: Starting the level count at 0 instead of 1.

Interview preparation tip

Master the "Queue with Size" BFS template. It is the most reliable way to process trees level by level. When asked for "Minimum" or "Maximum" at a level, this approach is less error-prone than DFS.