0807MediumTreeDepth-First SearchBinary Search TreeBinary Tree

Heavier Snowfall Totals

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Trains the technique from

LeetCode 1038Binary Search Tree to Greater Sum Tree

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A ski patrol files every snowfall it has measured, in centimetres, in a binary search tree keyed by the depth: for any filed fall, every depth in the branch to its left is smaller and every depth in the branch to its right is larger. No two filed falls share a depth.

The tree arrives as records, a level-order listing. The first entry is the depth at the head of the tree. After that, the listing gives the first branch then the second branch of each record already listed, in the order those records appear, writing null wherever a branch is empty. Slots below a null are never written down, and the trailing run of null entries is left off.

The patrol wants each record restated as a heavier-fall total: the depth of that fall plus the depths of every strictly greater fall on file. So a depth of 12 filed alongside depths 30 and 45 is restated as 12 + 30 + 45, and the greatest depth on file is restated as itself.

Rewrite every record with its heavier-fall total. The shape of the tree does not change, only the numbers written in it, and the rewritten numbers are generally no longer in search-tree order.

Return the rewritten tree in the same level-order form: null for an empty branch, nothing written below a null, and no trailing null entries.

Examples

Example 1

Input: records = [30, 12, 45, 6, 18, 40, 60]
Output: [175, 205, 105, 211, 193, 145, 60]

The filed depths are 6, 12, 18, 30, 40, 45 and 60, which add up to 211. The record at the head holds 30, and 30 with the heavier depths 40, 45 and 60 comes to 175, so that slot reports 175. The 60 has nothing heavier on file and reports 60, while the lightest depth 6 picks up everything and reports 211.

Example 2

Input: records = [7]
Output: [7]

One record is on file and nothing is heavier than it, so it reports its own depth of 7.

Example 3

Input: records = [1, null, 5, null, 6]
Output: [12, null, 11, null, 6]

Every record here hangs off the second branch of the one above it, giving depths 1, 5 and 6. The 6 reports 6, the 5 reports 5 + 6 = 11 and the 1 reports 1 + 5 + 6 = 12. The shape is untouched, so the two nulls stay exactly where they were.

Constraints

The number of records in the tree is in the range [1, 100].
1 <= records.length <= 200
0 <= records[i] <= 100
An entry of records that is null marks an empty branch rather than a record.
records[0] is never null, and no depth appears twice.
Every heavier-fall total therefore stays at or below 5050.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def rewrite_snow_records(records: list[int | None]) -> list[int | None]:

Java

public Integer[] rewriteSnowRecords(Integer[] records)

The intended approach

Time O(m) for a listing of m slots: one sweep to wire the tables, one ordered walk, one rebuild.Space O(m) for the index tables and the returned listing, plus O(h) for the stack on a tree of height h.

A queue sweep wires the listing into first-branch and second-branch index tables. The walk that descends every second branch, takes the node, then steps once down the first branch delivers the depths from greatest to smallest, so one running total is enough: add the record's own depth to the total and write the total back into that record. An explicit stack drives the walk, so a hundred-record chain is fine. Rebuilding the listing from the same index tables reproduces the original shape with the new numbers in it, and the trailing run of nulls comes off at the end.

Pythonentry point: rewrite_snow_records

def _decode(listing):
    """Compact level-order listing -> (values, left, right) index tables."""
    if not listing or listing[0] is None:
        return [], [], []
    left = [-1] * len(listing)
    right = [-1] * len(listing)
    queue = [0]
    head = 0
    slot = 1
    while head < len(queue) and slot < len(listing):
        here = queue[head]
        head += 1
        if slot < len(listing):
            if listing[slot] is not None:
                left[here] = slot
                queue.append(slot)
            slot += 1
        if slot < len(listing):
            if listing[slot] is not None:
                right[here] = slot
                queue.append(slot)
            slot += 1
    return list(listing), left, right


def _encode(values, left, right):
    """(values, left, right) index tables -> compact level-order listing."""
    if not values:
        return []
    out = [values[0]]
    order = [0]
    head = 0
    while head < len(order):
        here = order[head]
        head += 1
        for table in (left, right):
            kid = table[here]
            if kid == -1:
                out.append(None)
            else:
                out.append(values[kid])
                order.append(kid)
    while out and out[-1] is None:
        out.pop()
    return out

def rewrite_snow_records(records):
    # Walking the ordered tree the other way round -- second branch, node, first branch
    # -- hands the depths over from heaviest to lightest, so a single running total
    # carries exactly the depths already passed. Adding the current depth to it before
    # writing gives that record its heavier-fall total in one sweep, and the sweep runs
    # off an explicit stack so a hundred-record chain costs no interpreter recursion.
    # Time O(m) for a listing of m slots, space O(m) for the tables plus O(h) stack.
    values, left, right = _decode(records)
    running = 0
    stack = []
    here = 0 if values else -1
    while stack or here != -1:
        while here != -1:
            stack.append(here)
            here = right[here]
        node = stack.pop()
        running += values[node]
        values[node] = running
        here = left[node]
    return _encode(values, left, right)

The obvious version

Never wire a branch link at all. The filed depths are the non-null entries of the listing, and the rewrite leaves every record where it sits, so walk the listing slot by slot and replace each depth with the sum of every filed depth that is not smaller than it, copying the nulls straight across.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_rewrite_snow_records(records):
    # The rewrite never moves a record, so each slot of the listing can be recomputed
    # where it sits: add up every filed depth that is not lighter than this one.
    depths = [v for v in records if v is not None]
    out = []
    for slot in records:
        if slot is None:
            out.append(None)
            continue
        total = 0
        for d in depths:
            if d >= slot:
                total += d
        out.append(total)
    while out and out[-1] is None:
        out.pop()
    return out

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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