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2385. Amount of Time for Binary Tree to Be Infected

Description

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.

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# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def amountOfTime(self, root: Optional[TreeNode], start: int) -> int:
        def dfs(node: Optional[TreeNode], fa: Optional[TreeNode]):
            if node is None:
                return
            if fa:
                g[node.val].append(fa.val)
                g[fa.val].append(node.val)
            dfs(node.left, node)
            dfs(node.right, node)

        def dfs2(node: int, fa: int) -> int:
            ans = 0
            for nxt in g[node]:
                if nxt != fa:
                    ans = max(ans, 1 + dfs2(nxt, node))
            return ans

        g = defaultdict(list)
        dfs(root, None)
        return dfs2(start, -1)

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private Map<Integer, List<Integer>> g = new HashMap<>();

    public int amountOfTime(TreeNode root, int start) {
        dfs(root, null);
        return dfs2(start, -1);
    }

    private void dfs(TreeNode node, TreeNode fa) {
        if (node == null) {
            return;
        }
        if (fa != null) {
            g.computeIfAbsent(node.val, k -> new ArrayList<>()).add(fa.val);
            g.computeIfAbsent(fa.val, k -> new ArrayList<>()).add(node.val);
        }
        dfs(node.left, node);
        dfs(node.right, node);
    }

    private int dfs2(int node, int fa) {
        int ans = 0;
        for (int nxt : g.getOrDefault(node, List.of())) {
            if (nxt != fa) {
                ans = Math.max(ans, 1 + dfs2(nxt, node));
            }
        }
        return ans;
    }
}

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int amountOfTime(TreeNode* root, int start) {
        unordered_map<int, vector<int>> g;
        function<void(TreeNode*, TreeNode*)>