0632MediumTreeDepth-First SearchBreadth-First SearchBinary Tree

Sensors With A Clear View

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Trains the technique from

LeetCode 1448Count Good Nodes 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 survey network of altitude sensors is wired as a binary tree. tree is the level-order picture of the network. tree[0] is the root. After that the entries arrive in pairs: the next pair gives the left child and then the right child of the next node already read, and null marks a child that is not there. Pairs that would be all null at the end of the list may be left off. Each entry that is not null is the altitude of that sensor in metres, and altitudes may be negative.

Follow the links from the root down to a sensor. That sensor has a clear view when nothing on the way, counting the root and every sensor in between, stands strictly higher than it does. Return the number of sensors with a clear view.

Examples

Example 1

Input: tree = [4, 4, 2, null, 3, 1, 5]
Output: 3

The root sits at 4 and has a clear view. Its left child also sits at 4, and nothing on the way to it is above 4, so it counts. Its right child at 2 does not, nor does the sensor at 3 below the left child, nor the sensor at 1. The sensor at 5 has only 4 and 2 above it, so it counts. That makes 3.

Example 2

Input: tree = [-5, -6, -4]
Output: 2

The root at -5 counts. Its left child at -6 has the root above it at -5, which is higher, so it does not count. Its right child at -4 has only -5 above it, so it counts. That makes 2.

Example 3

Input: tree = [9, 3, 4, null, 5]
Output: 1

Only the root at 9 counts. The sensors at 3 and 4 sit below the root, and the sensor at 5, reached through the sensor at 3, still has the root at 9 on its path.

Constraints

The number of sensors in the tree is in the range [1, 10^5].
-10^4 <= tree[i] <= 10^4 for every entry that is not 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 good_nodes(tree: list) -> int:

Java

public int goodNodes(Integer[] tree)

The intended approach

Time O(n) — each sensor is decoded once and popped once.Space O(n) for the child arrays plus the stack, which a chain fills to the depth of the tree.

Rebuild the child links from the level-order list, then push the root onto a stack paired with its own altitude. Popping a pair gives a sensor and the highest altitude found on the way down to it; the sensor counts when its altitude is not below that figure. Each child is pushed with the larger of the two altitudes, so the figure travelling down each branch stays correct. The stack is explicit because the tree may be one long chain of 100000 sensors.

Pythonentry point: good_nodes

def _decode(tree):
    vals = [tree[0]]
    left = [-1]
    right = [-1]
    order = [0]
    head = 0
    i = 1
    n = len(tree)
    while head < len(order) and i < n:
        node = order[head]
        head += 1
        for side in (0, 1):
            if i >= n:
                break
            v = tree[i]
            i += 1
            if v is None:
                continue
            vals.append(v)
            left.append(-1)
            right.append(-1)
            child = len(vals) - 1
            if side == 0:
                left[node] = child
            else:
                right[node] = child
            order.append(child)
    return vals, left, right


def good_nodes(tree):
    # O(n) time, O(n) space. Decode the level-order list, then walk down from the
    # root with an explicit stack that carries, alongside each sensor, the
    # highest altitude seen on the way to it. A chain of 100000 sensors would
    # blow a recursive walk, so the stack is kept by hand. A sensor is counted
    # when its own altitude is at least that carried high-water mark, and the
    # mark handed to its children is the larger of the two.
    vals, left, right = _decode(tree)
    total = 0
    stack = [(0, vals[0])]
    while stack:
        node, best = stack.pop()
        v = vals[node]
        if v >= best:
            total += 1
        nb = best if best > v else v
        if left[node] != -1:
            stack.append((left[node], nb))
        if right[node] != -1:
            stack.append((right[node], nb))
    return total

The obvious version

Record each sensor's parent while decoding. Then for every sensor separately, climb link by link back to the root, keeping the largest altitude met, and test that sensor against it.

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 _bfdecode(tree):
    vals = [tree[0]]
    up = [-1]
    order = [0]
    head = 0
    i = 1
    n = len(tree)
    while head < len(order) and i < n:
        node = order[head]
        head += 1
        for _ in range(2):
            if i >= n:
                break
            v = tree[i]
            i += 1
            if v is None:
                continue
            vals.append(v)
            up.append(node)
            order.append(len(vals) - 1)
    return vals, up


def bf_good_nodes(tree):
    vals, up = _bfdecode(tree)
    total = 0
    for start in range(len(vals)):
        top = vals[start]
        cur = up[start]
        while cur != -1:
            if vals[cur] > top:
                top = vals[cur]
            cur = up[cur]
        if vals[start] >= top:
            total += 1
    return total

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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