What is the brute force approach and complexity for Amount of Time for Binary Tree to Be Infected?

For each node, count its edges to start; the answer is the largest count with time complexity O(N²) and space complexity O(H).

What is the optimal approach and complexity for Amount of Time for Binary Tree to Be Infected?

DFS records parents; BFS from start over left, right and parent; the answer is rounds − 1 with optimal time complexity O(N) and space complexity O(N).

How do you identify the pattern for Amount of Time for Binary Tree to Be Infected?

Look for key signals in the prompt: "infection spreads to neighbours each minute" indicates Each minute is one edge further. That is one BFS round, starting from the start node.; "neighbours include the parent" indicates In a tree the parent is a neighbour too. The tree must act as an undirected graph, so we need parent links.; "minutes until the whole tree is infected" indicates The answer is the distance to the farthest node, which is the last BFS layer..

Amount of Time for Binary Tree to Be Infected

Medium·LeetCode #2385 Pro OnlyInterview-Level

LeetCode

Problem Statement

You are given the root of a binary tree with unique values, and an integer start. At minute 0, an infection starts at the node whose value is start.

Each minute, every uninfected node that is next to an infected node becomes infected. A node's neighbours are its left child, its right child and its parent.

Return the number of minutes until every node is infected.

Input

The root of a binary tree with unique values, and the value start of the first infected node.

Output

The number of minutes until the whole tree is infected.

Example 1

Input: root = [1, 5, 3, null, 4, 10, 6, 9, 2], start = 3

Output: 4

Minute 0: node 3. Minute 1: nodes 1, 10 and 6. Minute 2: node 5. Minute 3: node 4. Minute 4: nodes 9 and 2. The whole tree is infected after 4 minutes.

Example 2

Input: root = [1], start = 1

Output: 0

The only node is infected at minute 0, so nothing else needs time.

Constraints

• The number of nodes is in the range [1, 10^5]
• 1 ≤ Node.val ≤ 10^5
• Each node has a unique value
• A node with the value `start` exists in the tree

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“infection spreads to neighbours each minute”

→ Each minute is one edge further. That is one BFS round, starting from the start node.

“neighbours include the parent”

→ In a tree the parent is a neighbour too. The tree must act as an undirected graph, so we need parent links.

“minutes until the whole tree is infected”

→ The answer is the distance to the farthest node, which is the last BFS layer.

Approach Complexity Ladder

Brute Force — distance from start to every nodeO(N²)O(H)

For each node, count its edges to start; the answer is the largest count

Optimal — parent map, then BFS from start, count the roundsO(N)O(N)

DFS records parents; BFS from start over left, right and parent; the answer is rounds − 1

Brute Force Intuition

A node gets infected at the minute equal to its distance from the start node. So find the distance of every node from start and return the largest. Each distance is found by climbing both root paths to where they split.

Recognition rule for this sub-pattern:

BFS reaches nodes in order of distance, so the first time you hit the target is the shortest. Add parent links to go upward.

Remember this · Tree BFS

Trigger

Level order, per-level max/average, right side view, minimum depth, nodes at distance k.

Think

Does the question care about depth / rows rather than paths?

Key question

How many nodes are in the current level (queue size at the start of the level)?

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