0756EasyTreeDepth-First SearchBreadth-First SearchBinary Tree

Average Reading Per Tree Tier

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Trains the technique from

LeetCode 637Average of Levels in Binary 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 monitoring rig is wired as a binary tree. Each station holds one integer reading and has at most two downstream stations, a left one and a right one.

The rig arrives as readings, a flat level-order encoding of that tree. Entry 0 is the root station's reading. After that, taking the present stations in level-order, each one claims the next two entries as its left child slot and then its right child slot; a slot holding null means that child is absent. Slots that would sit past the end of the list are absent too, so trailing null entries may be left off.

Group the stations by tier, where the root sits in tier 0 and a child sits one tier below its parent. Return the mean reading of each tier, ordered from tier 0 downwards. Each returned value is a decimal number and is accepted when it is within 1e-5 of the true mean.

Examples

Example 1

Input: readings = [4, 2, 9, 1, null, 6, 8]
Output: [4.0, 5.5, 5.0]

Tier 0 holds the single reading 4. Tier 0's two child slots give tier 1 the readings 2 and 9, whose mean is 5.5. Station 2 has a left child reading 1 and no right child, and station 9 has children reading 6 and 8, so tier 2 holds 1, 6 and 8 with mean 5.

Example 2

Input: readings = [6, 5, 4, -1, 0, 2]
Output: [6.0, 4.5, 0.3333333333333333]

Tier 1 holds 5 and 4. Station 5 claims entries 3 and 4, reading -1 and 0; station 4 claims entry 5, reading 2, and its right slot falls past the end of the list, so it is absent. Tier 2 therefore holds -1, 0 and 2, whose mean is one third.

Example 3

Input: readings = [7, 3, 12]
Output: [7.0, 7.5]

The root reads 7, so tier 0 has mean 7. Its two children read 3 and 12, so tier 1 has mean 7.5, and neither child claims any further entries.

Constraints

The rig holds between 1 and 10^4 stations.
1 <= readings.length <= 2 * 10^4 + 1
-2^31 <= readings[i] <= 2^31 - 1
readings[0] is a reading, never null.

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 average_per_tier(readings: list) -> list[float]:

Java

public double[] averagePerTier(Integer[] readings)

The intended approach

Time O(n) for an encoding of n entries: the cursor crosses the list once.Space O(w) for the widest tier, plus O(h) for the returned list of h tier means.

Because each present station claims exactly two consecutive entries for its child slots, the encoding can be walked one tier at a time with a single cursor: hold the current tier's readings, record their mean, then advance the cursor two entries per station in that tier and keep the non-null values as the next tier. No station objects are built.

Pythonentry point: average_per_tier

def average_per_tier(readings):
    # Walk the encoding tier by tier: every present node in a tier owns the next
    # two entries, so a single moving cursor is enough to find the next tier.
    # Time O(n), space O(w) for the widest tier.
    total = len(readings)
    tier = [readings[0]]
    cursor = 1
    out = []
    while tier:
        out.append(sum(tier) / len(tier))
        nxt = []
        for _ in tier:
            for _ in range(2):
                if cursor < total:
                    v = readings[cursor]
                    cursor += 1
                    if v is not None:
                        nxt.append(v)
        tier = nxt
    return out

The obvious version

Materialise the stations first, giving each one explicit left and right child links, then run a depth-first walk from an explicit stack that adds every reading into a per-depth running sum and count. The means come out of those tallies at the end.

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_average_per_tier(readings):
    total = len(readings)
    vals = [readings[0]]
    left = [None]
    right = [None]
    order = [0]
    cursor = 1
    head = 0
    while head < len(order):
        node = order[head]
        head += 1
        for slot in (0, 1):
            child = None
            if cursor < total:
                v = readings[cursor]
                cursor += 1
                if v is not None:
                    vals.append(v)
                    left.append(None)
                    right.append(None)
                    child = len(vals) - 1
                    order.append(child)
            if slot == 0:
                left[node] = child
            else:
                right[node] = child

    sums = {}
    counts = {}
    stack = [(0, 0)]
    while stack:
        node, depth = stack.pop()
        sums[depth] = sums.get(depth, 0) + vals[node]
        counts[depth] = counts.get(depth, 0) + 1
        if right[node] is not None:
            stack.append((right[node], depth + 1))
        if left[node] is not None:
            stack.append((left[node], depth + 1))
    return [sums[d] / counts[d] for d in range(len(sums))]

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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