Create 2385-amount-of-time-for-binary-tree-to-be-infected.java

Author: sujal-goswami

java/2385-amount-of-time-for-binary-tree-to-be-infected.java

@@ -0,0 +1,56 @@
+/*--------------------------
+  Time Complexity: O(n)
+  Space Complexity: O(n)
+---------------------------*/  
+class Solution {
+    public int amountOfTime(TreeNode root, int start) {
+        Map<Integer, List<Integer>> g = treeTograph(root);
+        int time = 0;
+        Queue<Integer> q = new LinkedList<>();
+        Set<Integer> visited = new HashSet<>();
+        q.add(start);
+        visited.add(start);
+
+        while(!q.isEmpty()){
+            int size = q.size(); 
+            for (int i = 0; i < size; i++) {
+                int curr = q.poll();
+                for (int neighbour : g.get(curr)) {
+                    if (!visited.contains(neighbour)) {
+                        q.add(neighbour);
+                        visited.add(neighbour);
+                    }
+                }
+            }
+            time++;
+        }
+        return time-1;
+    }
+    public HashMap<Integer, List<Integer>> treeTograph(TreeNode root) {
+        HashMap<Integer, List<Integer>> graph = new HashMap<>();
+        buildGraph(root, graph);
+        return graph;
+    }
+
+    private void buildGraph(TreeNode node, HashMap<Integer, List<Integer>> graph) {
+        if (node == null) {
+            return;
+        }
+
+        graph.putIfAbsent(node.val, new ArrayList<>());
+
+        if (node.left != null) {
+            graph.get(node.val).add(node.left.val);
+            graph.putIfAbsent(node.left.val, new ArrayList<>());
+            graph.get(node.left.val).add(node.val);
+            buildGraph(node.left, graph);
+        }
+
+        if (node.right != null) {
+            graph.get(node.val).add(node.right.val);
+            graph.putIfAbsent(node.right.val, new ArrayList<>());
+            graph.get(node.right.val).add(node.val);
+            buildGraph(node.right, graph);
+        }
+    }
+}