0179MediumTreeBreadth-First SearchBinary Tree

Relay Tower Snake Sweep

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Trains the technique from

LeetCode 103Binary Tree Zigzag 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 broadcast network is wired as a binary tree of relay towers. Every tower carries a signed drift reading, and each tower feeds at most two towers below it.

The network arrives as towers, a flat listing built depth by depth: towers[0] is the tower at the top, and after it come the two feeds of each listed tower, in the order those towers appear. A missing feed is written as null, and nothing is listed underneath a null. Trailing null entries are omitted. An empty listing means the network has no towers at all.

An inspector logs the network one depth at a time, but alternates the direction of travel: the topmost depth is logged from left to right, the depth under it from right to left, the one after that from left to right, and so on down the network.

Return a list whose i-th entry is the list of drift readings at depth i, in the order the inspector logs them. Return an empty list for a network with no towers.

Examples

Example 1

Input: towers = [4, 9, 3, null, null, 6, 8]
Output: [[4], [3, 9], [6, 8]]

Depth 0 holds 4. Depth 1 holds 9 then 3 in wiring order, logged in reverse. The tower holding 9 feeds nothing, so depth 2 is 6 then 8 and is logged forward.

Example 2

Input: towers = [-100, 100, 0, 5, null, null, -7]
Output: [[-100], [0, 100], [5, -7]]

Readings can sit at either end of the allowed range. Depth 1 reverses to 0 then 100, while depth 2 keeps the wiring order 5 then -7.

Example 3

Input: towers = [5, null, 3, null, 2]
Output: [[5], [3], [2]]

Each depth carries a single tower, so reversing a depth changes nothing.

Constraints

The number of towers is in the range [0, 2000].
-100 <= drift reading <= 100

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 snake_sweep(towers: list) -> list[list[int]]:

Java

public List<List<Integer>> snakeSweep(Integer[] towers)

The intended approach

Time O(n)Space O(n)

Consume the flat listing once to rebuild the tower tree, since the two feeds of every listed tower follow in the same order the towers were listed. Then hold one depth at a time as a frontier, emit it forward on even depths and reversed on odd ones, and replace it with the towers it feeds.

Pythonentry point: snake_sweep

def snake_sweep(towers):
    # Rebuild the tower tree from the listing, then walk it one depth at a time,
    # flipping the row on every odd depth.
    # Time O(n): each entry is consumed once while building and each tower is
    # visited once while sweeping. Space O(n) for the rebuilt tree plus the
    # widest frontier.
    if not towers or towers[0] is None:
        return []
    root = {"val": towers[0], "kids": []}
    pending = [root]
    read = 0
    i = 1
    while read < len(pending) and i < len(towers):
        node = pending[read]
        read += 1
        for _ in range(2):
            if i >= len(towers):
                break
            value = towers[i]
            i += 1
            if value is not None:
                child = {"val": value, "kids": []}
                node["kids"].append(child)
                pending.append(child)
    rows = []
    frontier = [root]
    depth = 0
    while frontier:
        row = [node["val"] for node in frontier]
        rows.append(row[::-1] if depth % 2 else row)
        nxt = []
        for node in frontier:
            for child in node["kids"]:
                nxt.append(child)
        frontier = nxt
        depth += 1
    return rows

The obvious version

Never builds a tree. Walks the listing in blocks: the top depth occupies one slot, and each following depth occupies exactly two slots per real tower found on the depth above. Reverses every other block.

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_snake_sweep(towers):
    if not towers or towers[0] is None:
        return []
    rows = []
    i = 0
    slots = 1
    depth = 0
    while slots > 0 and i < len(towers):
        row = []
        used = 0
        while used < slots and i < len(towers):
            entry = towers[i]
            i += 1
            used += 1
            if entry is not None:
                row.append(entry)
        if not row:
            break
        rows.append(row[::-1] if depth % 2 == 1 else row)
        slots = 2 * len(row)
        depth += 1
    return rows

How this problem was checked

Reference solution executed against 10 test cases — 10 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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