Traverse a tree and answer 2D prefix sums
Company: BlackRock
Role: Software Engineer
Category: Coding & Algorithms
Difficulty: easy
Interview Round: Technical Screen
Quick Answer: This question evaluates a candidate's competency in tree traversal and array preprocessing, specifically in-order traversal of binary trees and 2D prefix-sum techniques for answering submatrix-sum queries.
In-order Traversal of a Binary Tree
Constraints
- 0 <= number of nodes <= 10^5
- Node values fit in a 32-bit signed integer
- An empty list represents an empty tree (return an empty list)
- None / null entries in the level-order list mark absent nodes
Examples
Input: ([1, None, 2, 3],)
Expected Output: [1, 3, 2]
Explanation: Root 1 has only a right child 2; node 2's left child is 3. Left of 1 is empty, visit 1, then in 2's subtree go left to 3 first, then 2.
Input: ([],)
Expected Output: []
Explanation: Empty tree yields an empty traversal.
Input: ([1],)
Expected Output: [1]
Explanation: Single node: no left, visit it, no right.
Input: ([4, 2, 6, 1, 3, 5, 7],)
Expected Output: [1, 2, 3, 4, 5, 6, 7]
Explanation: A complete BST; in-order traversal of a BST yields sorted values.
Input: ([3, 1, 2],)
Expected Output: [1, 3, 2]
Explanation: Root 3, left child 1, right child 2: visit left (1), node (3), right (2).
Input: ([5, 3, 8, 1, 4, None, 9],)
Expected Output: [1, 3, 4, 5, 8, 9]
Explanation: Node 8 has no left child (None) and right child 9; in-order: 1,3,4 (left subtree), 5 (root), then 8, 9 (right subtree).
Hints
- In-order = left, node, right. A recursive solution is the most direct; an iterative one uses an explicit stack to avoid deep recursion on a 10^5-node skewed tree.
- First reconstruct the tree from the level-order list with a BFS queue: pop a parent, attach its next two (non-null) entries as left and right children.
- For the iterative traversal: push left children until you hit null, pop and record a node, then move to its right child and repeat.
2D Prefix Sum (Top-Left Submatrix Queries)
Constraints
- 1 <= m, n (matrix may also be empty -> all sums are 0)
- 0 <= x < m and 0 <= y < n for every query
- Matrix values may be negative
- Sums may exceed 32-bit range; use 64-bit integers
Examples
Input: ([[1, 2, 3], [4, 5, 6], [7, 8, 9]], [[0, 0], [1, 1], [2, 2], [2, 0], [0, 2]])
Expected Output: [1, 12, 45, 12, 6]
Explanation: sum(0,0)=1; sum(1,1)=1+2+4+5=12; sum(2,2)=whole matrix=45; sum(2,0)=1+4+7=12 (first column); sum(0,2)=1+2+3=6 (first row).
Input: ([[5]], [[0, 0]])
Expected Output: [5]
Explanation: Single-cell matrix; the only sum is the cell itself.
Input: ([[-1, -2], [-3, -4]], [[0, 0], [0, 1], [1, 0], [1, 1]])
Expected Output: [-1, -3, -4, -10]
Explanation: All-negative matrix: sum(0,0)=-1; sum(0,1)=-1-2=-3; sum(1,0)=-1-3=-4; sum(1,1)=-1-2-3-4=-10.
Input: ([[2, -1, 3], [0, 4, -2]], [[1, 2], [0, 1], [1, 0]])
Expected Output: [6, 1, 2]
Explanation: Mixed signs: sum(1,2)=2-1+3+0+4-2=6; sum(0,1)=2-1=1; sum(1,0)=2+0=2.
Input: ([[1000000000, 1000000000], [1000000000, 1000000000]], [[1, 1]])
Expected Output: [4000000000]
Explanation: Four cells of 1e9 sum to 4e9, which overflows 32-bit signed range, confirming 64-bit accumulation is required.
Input: ([], [])
Expected Output: []
Explanation: Empty matrix with no queries returns an empty list.
Hints
- Naively summing the rectangle for each query is O(m·n) per query — too slow when there are many queries.
- Define pre[i+1][j+1] = sum of the submatrix from (0,0) to (i,j). It satisfies pre[i+1][j+1] = A[i][j] + pre[i][j+1] + pre[i+1][j] - pre[i][j] (inclusion-exclusion).
- With that table, the answer to a top-left query (x, y) is simply pre[x+1][y+1]. The 1-based padding row/column avoids special-casing the first row and column.