0244MediumTreeBreadth-First SearchBinary Tree

Cable Splice Depths

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Trains the technique from

LeetCode 102Binary Tree Level Order Traversal

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 fibre trunk fans out through a set of splice enclosures. One enclosure sits at the head of the run, and every enclosure passes the fibre on to at most two enclosures further out: one wired to its left tray and one wired to its right tray. Each enclosure is labelled with a signed loss figure in hundredths of a decibel.

Because the harness passes plain JSON, the run arrives as the array splices, written out depth by depth. splices[0] is the loss figure at the head enclosure. After it the entries come in pairs, giving the left tray and then the right tray of each enclosure already written out, taken in that same order. An empty tray is written null, and a null claims no pair of its own. Trailing null entries are left off the end, and an empty array means no enclosures were fitted.

The depth of the head enclosure is 0, and an enclosure wired to a tray of a depth-d enclosure sits at depth d + 1.

Return a list whose i-th entry is the list of loss figures of every enclosure at depth i, each depth given in the order those enclosures appear in splices. Return an empty list when no enclosures were fitted.

Examples

Example 1

Input: splices = [4, -7, 12, null, 5]
Output: [[4], [-7, 12], [5]]

The head enclosure reads 4. Its two trays carry -7 and 12, so depth 1 holds both figures in that order. The left tray of the -7 enclosure is empty and its right tray carries 5, the only enclosure at depth 2.

Example 2

Input: splices = [0]
Output: [[0]]

One enclosure was fitted and it reads 0, so depth 0 is the only depth reported.

Example 3

Input: splices = [-3, null, 8, 6, -1]
Output: [[-3], [8], [6, -1]]

The head enclosure has an empty left tray, so the entry 8 belongs to its right tray. The 8 enclosure is then the next one written out, and the pair 6 and -1 fills its two trays at depth 2.

Constraints

0 <= number of enclosures <= 2000
-1000 <= loss figure <= 1000
0 <= splices.length <= 4001
Every entry of splices is either null or an integer loss figure
splices[0] is a loss figure whenever splices is non-empty

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 splice_depths(splices: list) -> list[list[int]]:

Java

public List<List<Integer>> spliceDepths(Integer[] splices)

The intended approach

Time O(n) for n enclosures, since each listing entry is consumed once and each enclosure is visited once.Space O(n) for the rebuilt enclosures, the widest frontier and the returned rows.

Consume the listing once to rebuild the enclosures, which works because the two trays of every listed enclosure follow in the same order those enclosures were listed. Then hold a single depth as a frontier: record the figures it carries, then replace it with every enclosure its trays feed, and stop when a depth feeds nothing.

Pythonentry point: splice_depths

def splice_depths(splices):
    if not splices or splices[0] is None:
        return []
    # Rebuild the enclosure tree: the two trays of every listed enclosure
    # follow in the listing in the same order the enclosures were listed.
    values = [splices[0]]
    trays = [[]]
    listed = 0
    at = 1
    while listed < len(values) and at < len(splices):
        for _ in range(2):
            if at >= len(splices):
                break
            entry = splices[at]
            at += 1
            if entry is not None:
                values.append(entry)
                trays.append([])
                trays[listed].append(len(values) - 1)
        listed += 1
    # Hold one depth at a time and replace it with the depth it feeds.
    rows = []
    frontier = [0]
    while frontier:
        rows.append([values[i] for i in frontier])
        below = []
        for i in frontier:
            below.extend(trays[i])
        frontier = below
    return rows

The obvious version

Rebuild the run keeping only a pointer from each enclosure to the one feeding it, then work out every enclosure's depth by walking that chain back to the head and drop the figure into that depth's bucket. O(n * h) for height h.

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_splice_depths(splices):
    if not splices or splices[0] is None:
        return []
    values = [splices[0]]
    above = [-1]
    listed = 0
    at = 1
    while listed < len(values) and at < len(splices):
        for _ in range(2):
            if at >= len(splices):
                break
            entry = splices[at]
            at += 1
            if entry is not None:
                values.append(entry)
                above.append(listed)
        listed += 1
    rows = []
    for node in range(len(values)):
        depth = 0
        walk = node
        while above[walk] != -1:
            walk = above[walk]
            depth += 1
        while len(rows) <= depth:
            rows.append([])
        rows[depth].append(values[node])
    return rows

How this problem was checked

Reference solution executed against 11 test cases — 11 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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