0780MediumTreeDepth-First SearchBinary TreeDP on Trees

Longest Stepped Run In The Pipe Tree

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Trains the technique from

LeetCode 549Binary Tree Longest Consecutive Sequence II

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 pressure gauge sits at every junction of a binary tree of pipes, and each junction has at most two junctions hanging below it.

The tree is handed to you as a level-order list readings. readings[0] is the root's gauge. Walk the list left to right while keeping a queue of the junctions recorded so far, starting with the root: the next two entries are the lower-left and lower-right slots of the first queued junction, the two after that are the slots of the second queued junction, and so on. An empty slot is written as null, and nothing at all is written for the slots below a null, so a null takes exactly one entry of the list. Once every remaining entry would be null the list may simply stop.

A route is a sequence of one or more distinct junctions in which each junction is joined by a pipe to the next one, going either down to a junction below it or up to the junction above it. A route may therefore climb from one junction up towards the root and then descend into a different branch, and it turns at most once because a tree offers no other way through.

A route is stepped when the gauge changes by exactly one at every step and always in the same direction, so the readings along it either climb by one each time or drop by one each time. Return the number of junctions on the longest stepped route in the tree; a single junction on its own counts as a stepped route of length 1.

Examples

Example 1

Input: readings = [4,5,3,6,null,null,2]
Output: 5

The root reads 4, its lower-left junction 5 with a 6 below that, and its lower-right junction 3 with a 2 below that. The route 6, 5, 4, 3, 2 climbs from the 6 up to the root and then descends the other branch, dropping by one at every step, so it is a stepped route through 5 junctions.

Example 2

Input: readings = [5,6,null,5]
Output: 2

The root reads 5, its lower-left junction 6, and below that a 5 again. The whole route 5, 6, 5 changes by one at each step but reverses direction in the middle, so it is not stepped; the pair 5, 6 is, giving 2.

Example 3

Input: readings = [3,3,3]
Output: 1

Every gauge reads the same, so no step of exactly one exists anywhere and only a single junction on its own qualifies.

Constraints

The tree holds k junctions with 1 <= k <= 3 * 10^4.
1 <= readings.length <= 6 * 10^4 + 1
-30000 <= readings[i] <= 30000
Every entry of `readings` is either a gauge reading or `null`, and `readings[0]` is a reading.

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 longest_consecutive(readings: list) -> int:

Java

public int longestConsecutive(Integer[] readings)

The intended approach

Time O(k) for k junctions: one pass to rebuild the tree and one reverse pass over the ids.Space O(k) for the rebuilt tree and the two arrays.

Rebuild the tree into parallel value / lower-left / lower-right arrays with one pass over the list, handing out ids in level order so every child id exceeds its parent's. Sweeping the ids backwards is then a bottom-up order, which lets two numbers be kept per junction: the longest run descending from it that climbs by one at each step, and the longest that drops by one. A run that turns at a junction uses a climbing run through one child and a dropping run through the other, so their lengths add with the junction counted once. No recursion is involved, so a tree that is one long chain is no problem.

Pythonentry point: longest_consecutive

def _rebuild(readings):
    """Level-order list -> parallel arrays (val, left, right), ids in level order."""
    val = [readings[0]]
    left = [-1]
    right = [-1]
    queue = [0]
    seat = 0
    at = 1
    while seat < len(queue) and at < len(readings):
        cur = queue[seat]
        seat += 1
        for slot in (0, 1):
            if at >= len(readings):
                break
            entry = readings[at]
            at += 1
            if entry is None:
                continue
            idx = len(val)
            val.append(entry)
            left.append(-1)
            right.append(-1)
            if slot == 0:
                left[cur] = idx
            else:
                right[cur] = idx
            queue.append(idx)
    return val, left, right


def longest_consecutive(readings):
    # Because _rebuild hands out ids in level order, every child id is larger than
    # its parent's, so one reverse sweep over the ids is already a bottom-up order
    # and no recursion or stack is needed.
    # rise[i] = longest run that walks down from i adding one at every step,
    # fall[i] = the same walking down subtracting one. A run that bends at i takes
    # one of each, from the two different children.
    # Time O(k), space O(k) for k nodes.
    val, left, right = _rebuild(readings)
    k = len(val)
    rise = [1] * k
    fall = [1] * k
    best = 1
    for i in range(k - 1, -1, -1):
        lr = lf = rr = rf = 0
        c = left[i]
        if c >= 0:
            if val[c] == val[i] + 1:
                lr = rise[c]
            elif val[c] == val[i] - 1:
                lf = fall[c]
        c = right[i]
        if c >= 0:
            if val[c] == val[i] + 1:
                rr = rise[c]
            elif val[c] == val[i] - 1:
                rf = fall[c]
        rise[i] = 1 + (lr if lr > rr else rr)
        fall[i] = 1 + (lf if lf > rf else rf)
        bend = lr + rf
        other = lf + rr
        if other > bend:
            bend = other
        if 1 + bend > best:
            best = 1 + bend
    return best

The obvious version

Decode the list by slicing off one whole level at a time into an undirected neighbour list, then start a walk at every junction that may only step to a neighbour reading exactly one more, never stepping back where it came from. The longest such walk is the answer, since any stepped route read in its climbing direction is one of them. Quadratic in the number of junctions.

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 _decode_by_levels(readings):
    """Second, independent decoder: slice the list one level at a time.

    Level 0 is the root entry. Each following level occupies exactly two entries
    per non-null node of the level above, in that level's left-to-right order, so
    the levels can be cut off the list without any queue bookkeeping. Returns the
    values and an undirected neighbour list.
    """
    vals = [readings[0]]
    near = [[]]
    row = [0]
    at = 1
    while row and at < len(readings):
        want = 2 * len(row)
        chunk = list(readings[at:at + want])
        while len(chunk) < want:
            chunk.append(None)
        at += want
        nxt = []
        for j, node in enumerate(row):
            for entry in (chunk[2 * j], chunk[2 * j + 1]):
                if entry is None:
                    continue
                idx = len(vals)
                vals.append(entry)
                near.append([])
                near[node].append(idx)
                near[idx].append(node)
                nxt.append(idx)
        row = nxt
    return vals, near


def bf_longest_consecutive(readings):
    vals, near = _decode_by_levels(readings)
    best = 1
    for start in range(len(vals)):
        stack = [(start, -1, 1)]
        while stack:
            node, came, run = stack.pop()
            if run > best:
                best = run
            for nb in near[node]:
                if nb != came and vals[nb] == vals[node] + 1:
                    stack.append((nb, node, run + 1))
    return best

How this problem was checked

Reference solution executed against 16 test cases — 16 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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