2385. Amount of Time for Binary Tree to Be Infected | Leetcode Solution


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.



  • 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.

The solution to this question is straightforward. If you know graphs, you can answer this question in less than a minute. We can app DFS and can get the answer, but how we can do that as this is a tree? Either we can make a new graph with this tree or we can traverse the tree once and then store the parent node of all the nodes. Then we can simply apply DFS on this.

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  1. #include
    using namespace std;

    int main(){

    cout<<"Welcome to FCFS \n";
    cout<<"Enter the number of Processes: ";
    int k=1;
    int pid,arrivalTime,burstTime;
    cout<<"Enter Process ID: ";
    cout<<"Enter Arrival Time: ";
    cout<<"Enter the Burst Time: ";
    int st=0;
    cout<<"Process No: Arrival Time: Burst Time: Completion Time: Turn Around Time: Waiting Time: Response Time:\n";
    for (int i = 0; i < process.size(); i++)
    int ct=max(v[1]+v[2],st+v[2]);
    cout<<v[0]<<" "<<v[1]<<" "<<v[2]<<" "<<ct<<" "<<ct-v[1]<<" "<<abs(st-v[1])<<" "<<max(0,st-v[1])<<endl;

    return 0;


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