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# 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 = , 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)])

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:
queue.append((nei.val, time + 1))

return max_infection``````
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