Transform tree with subtree sum mapping

You are given the roots of two complete binary trees T1 and T2 that have identical structure (same shape and number of nodes). Modify T2 so that, for every node position, the value stored at that node in T2 equals the sum of all node values in the corresponding subtree of T1 (including the root of that subtree). Implement transform(TreeNode t1, TreeNode t 2) to perform this in O(n) time, and state your space complexity. Describe the traversal order you use and how you map positions between the two trees. For testing, also implement buildCompleteTree(int[] levelOrder) that constructs a complete binary tree from a level-order array representation.

What a Strong Answer Covers

• A clear algorithm with the right data structures and enough pseudocode or code-level detail to implement it.
• A correctness argument that explains why the algorithm covers all required cases.
• Time and space complexity, plus at least one alternative approach when relevant.
• Focused tests for normal cases, edge cases, and failure modes.

Follow-up Questions

• How would the approach change if the input were streaming or too large for memory?
• What invariants would you assert in production code?
• Which tests would catch off-by-one, duplicate, or tie-breaking bugs?

Transform Tree With Subtree Sum Mapping

You are given two complete binary trees, T1 and T2, that have identical structure (same shape and number of nodes). For testing convenience the trees are passed in level-order array form: index 0 is the root, and the node at index i has its left child at 2i + 1 and its right child at 2i + 2.

Produce the new values for T2 so that, for every node position, the value equals the sum of all node values in the corresponding subtree of T1 (including that subtree's root). Implement transform(t1, t2) to return the transformed level-order array for T2 in O(n) time, and be ready to state the space complexity and traversal order you use.

The array layout encodes a buildCompleteTree(levelOrder)-style representation: because the tree is complete and the two trees share structure, position i in T1 maps directly to position i in T2. Process indices from last to first so each parent can read its already-computed children's subtree sums.

Example: T1 = [1, 2, 3, 4, 5, 6, 7], T2 = [10, 20, 30, 40, 50, 60, 70] -> [28, 11, 16, 4, 5, 6, 7] (root subtree sums to 28, left subtree of T1 sums to 2+4+5=11, right subtree sums to 3+6+7=16, leaves equal themselves).

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 2i + 1 and 2i + 2 (level-order complete-tree layout)

Hints

1. 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.
2. 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.
3. 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.

Solution

Solution Alignment

The prompt asks for an implementation-level answer. The safest way to present it is to define the state, maintain clear invariants, then walk through complexity and tests.

Problem Restatement

You are given the roots of two complete binary trees T1 and T2 that have identical structure (same shape and number of nodes). Modify T2 so that, for every node position, the value stored at that node in T2 equals the sum of all node values in the corresponding subtree of T1 (including the root of that subtree). Implement transform(TreeNode t1, TreeNode t 2) to perform this in O(n) time, and state your space complexity. Describe the traversal order you use and how you map positions between the two trees. For testing, also implement buildCompleteTree(int[] levelOrder) that constructs a complete binary tree from a level-order array representation.

Recommended Approach

Choose traversal based on the required view or aggregate. DFS is natural for subtree computations and reconstruction; BFS is natural for level order or side views. Keep per-depth or per-position state when the output depends on columns, rows, or depths.

Correctness

The implementation should maintain an invariant after each loop or operation that directly matches the problem statement. At termination, that invariant implies the returned value has considered every valid candidate exactly once, or has preserved the required data-structure state after every API call.

Complexity

Most tree traversals are O(n) time and O(h) recursion stack for DFS or O(w) queue space for BFS, where h is height and w is maximum width.

Edge Cases and Tests

Empty tree, one node, skewed tree, duplicate values when reconstruction assumes uniqueness, deep recursion, and tie-breaking for same row/column nodes.