Amount of Time for Binary Tree to Be Infected LeetCode Solution

Problem – 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.

Example 1:

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

Amount of Time for Binary Tree to Be Infected LeetCode Solution in C++

unordered_map<int,list<int>> v;  //adjacency list

//create undirected graph for every parent-child  -> BFS
void createGraph(TreeNode* root){
    queue<pair<TreeNode*,int>> q;
    q.push({root,-1});
    while(q.size()){
        auto [node,parent]= q.front(); q.pop();
        if(parent!=-1){
            v[parent].push_front(node->val);
            v[node->val].push_front(parent);
        }
        if(node->left)  q.push({node->left,node->val});
        if(node->right) q.push({node->right,node->val});
    }   
}

int amountOfTime(TreeNode* root, int start) {
    //create graph of given tree
    createGraph(root);

    //start bfs
    queue<int> q;
    unordered_map<int,bool> seen;
    q.push(start);
    seen[start]=true;
    int time=0;
    for(;q.size();time++){
        int n= q.size();
        while(n--){
            auto node= q.front();  q.pop();
            for(auto i:v[node]){
                if(!seen[i]){
                    q.push(i);
                    seen[i]=true;
                }
            }
        }
    }
    return time-1;
}

Amount of Time for Binary Tree to Be Infected LeetCode Solution in Java

class Solution {
    public int amountOfTime(TreeNode root, int start) {
        HashMap<TreeNode, TreeNode> mpp=new HashMap<>();
        TreeNode target=bfsToMapParents(root,mpp,start);
        
       return findMaxDistance(mpp, target);
    }
    private static int findMaxDistance(HashMap<TreeNode, TreeNode> mpp, TreeNode target) {
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(target);
        HashMap<TreeNode,Integer> vis = new HashMap<>();
        vis.put(target, 1);
        int maxi = 0;
        
        while(!q.isEmpty()) {
            int sz = q.size();
            int fl = 0;
            
            for(int i = 0;i<sz;i++) {
                TreeNode node = q.poll();
                if(node.left != null && vis.get(node.left) == null) {
                    fl = 1;
                    vis.put(node.left, 1);
                    q.offer(node.left);
                }
                if(node.right != null && vis.get(node.right) == null) {
                    fl = 1;
                    vis.put(node.right, 1);
                    q.offer(node.right);
                }

                if(mpp.get(node) != null && vis.get(mpp.get(node)) == null) {
                    fl = 1;
                    vis.put(mpp.get(node), 1);
                    q.offer(mpp.get(node));
                }
            }
            if(fl == 1) maxi++;
        }
        return maxi;
    }
     TreeNode bfsToMapParents(TreeNode root,HashMap<TreeNode, TreeNode> mpp, int start) {
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        TreeNode res = new TreeNode(-1);
        while(!q.isEmpty()) {
            TreeNode node = q.poll();
            if(node.val == start) res = node;
            if(node.left != null) {
                mpp.put(node.left, node);
                q.offer(node.left);
            }
            if(node.right != null) {
                mpp.put(node.right, node);
                q.offer(node.right);
            }
        }
        return res;
    }
}

Amount of Time for Binary Tree to Be Infected LeetCode Solution in Python

class Solution:
    def amountOfTime(self, root: Optional[TreeNode], start: int) -> int:
        graph = defaultdict(list)
        
        def build_graph(parent, node):
            if not node: return 
            
            if parent:
                graph[parent.val].append(node)
                graph[node.val].append(parent)
            
            build_graph(node, node.left)
            build_graph(node, node.right)
        
        build_graph(None, root)
        
        vis = set()
        max_infection = 0
        queue = deque([(start, 0)])
        vis.add(start)
        
        while queue:
            node_val, time = queue.popleft()
            max_infection = max(max_infection, time)
            
            for nei in graph[node_val]:
                if nei.val not in vis:
                    vis.add(nei.val)
                    queue.append((nei.val, time + 1))
        
        return max_infection

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