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Problem solution · Java

Amount of Time for Binary Tree to Be Infected

Amount of Time for Binary Tree to Be Infected: a Java solution using breadth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Breadth-first search
Source
walkccc LeetCode Solutions
Length
49 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Breadth-first search

For Amount of Time for Binary Tree to Be Infected, the implementation explores reachable states in layers, which is the standard shape for unweighted shortest paths and minimum-step transitions.

  1. Model each valid configuration as a state and each legal move as an edge.
  2. Seed the queue with the starting state and mark it immediately.
  3. Expand each state once, recording distance or reachability for unseen neighbours.

Code notes

  • 49 lines of Java from the credited upstream file 2385.java.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup, work queue.
  • 4 loop blocks detected.

Complexity

Verify that each state and transition is processed only a bounded number of times; that determines the traversal cost.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeAmount of Time for Binary Tree to Be Infected · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int amountOfTime(TreeNode root, int start) {    int ans = -1;    Map<Integer, List<Integer>> graph = getGraph(root);    Queue<Integer> q = new ArrayDeque<>(List.of(start));    Set<Integer> seen = new HashSet<>(Arrays.asList(start));     for (; !q.isEmpty(); ++ans) {      for (int sz = q.size(); sz > 0; --sz) {        final int u = q.poll();        if (!graph.containsKey(u))          continue;        for (final int v : graph.get(u)) {          if (seen.contains(v))            continue;          q.offer(v);          seen.add(v);        }      }    }     return ans;  }   private Map<Integer, List<Integer>> getGraph(TreeNode root) {    Map<Integer, List<Integer>> graph = new HashMap<>();    // (node, parent)    Queue<Pair<TreeNode, Integer>> q = new ArrayDeque<>(List.of(new Pair<>(root, -1)));     while (!q.isEmpty()) {      Pair<TreeNode, Integer> pair = q.poll();      TreeNode node = pair.getKey();      final int parent = pair.getValue();      if (parent != -1) {        graph.putIfAbsent(parent, new ArrayList<>());        graph.putIfAbsent(node.val, new ArrayList<>());        graph.get(parent).add(node.val);        graph.get(node.val).add(parent);      }      if (node.left != null)        q.add(new Pair<>(node.left, node.val));      if (node.right != null)        q.add(new Pair<>(node.right, node.val));    }     return graph;  }}

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