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.
Solutions
Solution 1: Two DFS
First, we build a graph through one DFS, and get an adjacency list $g$, where $g[node]$ represents all nodes connected to the node $node$.
Then, we use $start$ as the starting point, and search the entire tree through DFS to find the farthest distance, which is the answer.
The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the number of nodes in the binary tree.
/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */funcamountOfTime(root*TreeNode,startint)int{g:=map[int][]int{}vardfsfunc(*TreeNode,*TreeNode)dfs=func(node,fa*TreeNode){ifnode==nil{return}iffa!=nil{g[node.Val]=append(g[node.Val],fa.Val)g[fa.Val]=append(g[fa.Val],node.Val)}dfs(node.Left,node)dfs(node.Right,node)}vardfs2func(int,int)intdfs2=func(node,faint)(ansint){for_,nxt:=rangeg[node]{ifnxt!=fa{ans=max(ans,1+dfs2(nxt,node))}}return}dfs(root,nil)returndfs2(start,-1)}