Transform tree with subtree sum mapping
Company: Snowflake
Role: Software Engineer
Category: Coding & Algorithms
Difficulty: Medium
Interview Round: Technical Screen
Quick Answer: Transform tree with subtree sum mapping evaluates algorithm design, data structures, correctness, complexity, edge cases, and implementation details in a realistic interview setting. A strong answer states assumptions, handles edge cases, explains trade-offs, and shows how to validate the result clearly.
Constraints
- len(t1) == len(t2) (the two trees share identical structure)
- 0 <= n <= 10^5 nodes
- Node values fit in a signed 32-bit / 64-bit integer; values may be negative
- Children of index i are at 2*i + 1 and 2*i + 2 (level-order complete-tree layout)
Examples
Input: ([1, 2, 3, 4, 5, 6, 7], [10, 20, 30, 40, 50, 60, 70])
Expected Output: [28, 11, 16, 4, 5, 6, 7]
Explanation: Full 7-node tree: root = 1+2+3+4+5+6+7 = 28; left subtree = 2+4+5 = 11; right subtree = 3+6+7 = 16; the four leaves equal themselves.
Input: ([5], [99])
Expected Output: [5]
Explanation: Single node: its subtree is just itself, so the value becomes 5.
Input: ([], [])
Expected Output: []
Explanation: Empty tree edge case returns an empty array.
Input: ([1, 2, 3], [0, 0, 0])
Expected Output: [6, 2, 3]
Explanation: Root subtree sums to 1+2+3 = 6; the two leaves equal themselves.
Input: ([-1, -2, -3, -4, -5], [9, 9, 9, 9, 9])
Expected Output: [-15, -11, -3, -4, -5]
Explanation: Incomplete last level (5 nodes): index 1 has children 3 and 4, so subtree(1) = -2 + -4 + -5 = -11; index 2 has no children, so subtree(2) = -3; root = -1 + -11 + -3 = -15. Confirms negative values and a non-full bottom level are handled.
Input: ([10, 20], [1, 1])
Expected Output: [30, 20]
Explanation: Two nodes: index 1 is the root's left child with no children, so it stays 20; root = 10 + 20 = 30.
Hints
- Process node indices from the last position to the first. Because children always have a higher index than their parent in level-order, every child's subtree sum is finalized before its parent is reached.
- subtree_sum[i] = t1[i] + (subtree_sum[2i+1] if it exists) + (subtree_sum[2i+2] if it exists). Guard the child indices against n.
- Position i in T1 maps to position i in T2 because the trees are structurally identical, so you can write the result back to the same index. This is just a post-order traversal expressed over the array.