infectedBinaryTree.txt

2385. Amount of Time for Binary Tree to Be Infected

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

Each minute, a node becomes infected if:

The node is currently uninfected.
The node is adjacent to an infected node.
Return the number of minutes needed for the entire tree to be infected.

class Solution {
public:
    unordered_map<int, vector<int>> graph;

    int amountOfTime(TreeNode* root, int start) {
        constructGraph(root);
        queue<int> q;
        q.push(start);
        unordered_set<int> visited;
        int minutesPassed = -1;

        while (!q.empty()) {
            ++minutesPassed;
            for (int levelSize = q.size(); levelSize > 0; --levelSize) {
                int currentNode = q.front();
                q.pop();
                visited.insert(currentNode);
                for (int adjacentNode : graph[currentNode]) {
                    if (!visited.count(adjacentNode)) {
                        q.push(adjacentNode);
                    }
                }
            }
        }

        return minutesPassed;
    }

    void constructGraph(TreeNode* root) {
        if (!root) return;
        if (root->left) {
            graph[root->val].push_back(root->left->val);
            graph[root->left->val].push_back(root->val);
        }
        if (root->right) {
            graph[root->val].push_back(root->right->val);
            graph[root->right->val].push_back(root->val);
        }

        constructGraph(root->left);
        constructGraph(root->right);
    }
};