This problem requires converting a binary tree into an undirected graph and using either BFS or DFS to track the infection spread. The time complexity is linear in relation to the number of nodes. Efficient graph traversal with state tracking ensures the entire tree gets infected in the shortest time possible.
Problem Statement
You are given a binary tree represented by its root node and an integer start, which denotes the initially infected node. The infection starts at minute 0 from the start node.
At each minute, an infected node will cause its directly connected neighbors to become infected. The task is to determine how many minutes it will take for all the nodes in the tree to be infected.
Examples
Example 1
Input: root = [1,5,3,null,4,10,6,9,2], start = 3
Output: 4
The following nodes are infected during:
- 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 It takes 4 minutes for the whole tree to be infected so we return 4.
Example 2
Input: root = [1], start = 1
Output: 0
At minute 0, the only node in the tree is infected so we return 0.
Constraints
- The number of nodes in the tree is in the range [1, 105].
- 1 <= Node.val <= 105
- Each node has a unique value.
- A node with a value of start exists in the tree.
Solution Approach
Convert the Tree to an Undirected Graph
To solve this problem efficiently, convert the binary tree into an undirected graph where each node is connected to its parent and children. This transformation allows easy traversal using BFS or DFS, making it simpler to track the infection spread.
Breadth-First Search (BFS) for Infection Spread
Use BFS to simulate the infection spread across the graph. Start from the start node and expand to adjacent nodes level by level. Each level corresponds to a minute in the infection process. BFS ensures the shortest time for the infection to reach every node.
Track Infection with a State
Keep track of the nodes' infection states using a set or dictionary to avoid re-visiting nodes. This state tracking ensures that each node is infected only once, preventing redundant operations and ensuring efficiency.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | O(n) |
| Space | O(n) |
The time complexity is O(n) where n is the number of nodes in the tree, since BFS or DFS will visit each node exactly once. The space complexity is also O(n) due to the storage of the adjacency list (graph representation) and the state tracking structures used during traversal.
What Interviewers Usually Probe
- The problem tests the ability to convert a tree into an undirected graph.
- The candidate demonstrates understanding of BFS/DFS and state management in tree-like structures.
- The solution should involve careful handling of state to avoid redundant infections.
Common Pitfalls or Variants
Common pitfalls
- Failing to convert the tree into an undirected graph before starting traversal can lead to incorrect infection spread handling.
- Not keeping track of the visited nodes can cause infinite loops or re-infection of already infected nodes.
- Misunderstanding the BFS/DFS approach can lead to inefficient or incorrect spread simulation.
Follow-up variants
- The tree can be large with up to 105 nodes, requiring careful space and time optimization.
- The infection might not be continuous; ensure each minute correctly tracks the new infections.
- Edge cases where the tree consists of only one node or the infection starts at a leaf node should be handled.
How GhostInterview Helps
- GhostInterview helps break down the tree-to-graph conversion step, making the BFS/DFS traversal easier to implement.
- It clarifies how state tracking is crucial to avoid redundant infections, improving efficiency.
- The solver assists in optimizing space and time complexities, guiding you through performance considerations for large trees.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is the core pattern used to solve the Amount of Time for Binary Tree to Be Infected problem?
The core pattern is converting the binary tree into an undirected graph, followed by BFS or DFS for efficient infection spread simulation.
How can I optimize the BFS approach in this problem?
Optimizing BFS involves using a state tracking system to avoid revisiting infected nodes and ensuring the graph representation is accurate and efficient.
What are some common pitfalls when solving this problem?
Common pitfalls include failing to convert the tree into an undirected graph, not tracking visited nodes, and inefficient traversal techniques.
How does GhostInterview help with tree-based traversal problems?
GhostInterview helps by providing step-by-step guidance on graph conversions, traversal methods, and performance optimization, ensuring a clear and efficient approach.
Is BFS or DFS better for the Amount of Time for Binary Tree to Be Infected problem?
Both BFS and DFS are valid approaches, but BFS is typically preferred for this problem as it efficiently handles level-by-level infection spread.
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