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Formatted question description: https://leetcode.ca/all/2385.html

# 2385. Amount of Time for Binary Tree to Be Infected

• Difficulty: Medium.
• Related Topics: Tree, Depth-First Search, Breadth-First Search, Binary Tree.
• Similar Questions: Maximum Depth of Binary Tree, Shortest Path to Get Food, All Nodes Distance K in Binary Tree.

## Problem

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.

## Solution (Java, C++, Python)

• /**
* 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 int max = 0;

public int amountOfTime(TreeNode root, int start) {
dfs(root, start, new Distance(-1));
return max;
}

private int dfs(TreeNode root, int start, Distance l) {
if (root == null) {
return 0;
}
Distance ld = new Distance(-1);
Distance rd = new Distance(-1);
int left = dfs(root.left, start, ld);
int right = dfs(root.right, start, rd);
if (l.val == -1 && start == root.val) {
max = Math.max(left, right);
l.val = 1;
}
if (ld.val != -1) {
max = Math.max(max, ld.val + right);
l.val = ld.val + 1;
} else if (rd.val != -1) {
max = Math.max(max, rd.val + left);
l.val = rd.val + 1;
}
return Math.max(left, right) + 1;
}

private static class Distance {
int val;

Distance(int v) {
this.val = v;
}
}
}

############

/**
* 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);
Deque<Integer> q = new ArrayDeque<>();
Set<Integer> vis = new HashSet<>();
q.offer(start);
int ans = -1;
while (!q.isEmpty()) {
++ans;
for (int n = q.size(); n > 0; --n) {
int i = q.pollFirst();
if (g.containsKey(i)) {
for (int j : g.get(i)) {
if (!vis.contains(j)) {
q.offer(j);
}
}
}
}
}
return ans;
}

private void dfs(TreeNode root) {
if (root == null) {
return;
}
if (root.left != null) {
}
if (root.right != null) {
}
dfs(root.left);
dfs(root.right);
}
}

• # 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(root):
if root is None:
return
if root.left:
g[root.val].append(root.left.val)
g[root.left.val].append(root.val)
if root.right:
g[root.val].append(root.right.val)
g[root.right.val].append(root.val)
dfs(root.left)
dfs(root.right)

g = defaultdict(list)
dfs(root)
vis = set()
q = deque([start])
ans = -1
while q:
ans += 1
for _ in range(len(q)):
i = q.popleft()
for j in g[i]:
if j not in vis:
q.append(j)
return ans

############

# 2385. Amount of Time for Binary Tree to Be Infected
# https://leetcode.com/problems/amount-of-time-for-binary-tree-to-be-infected/

# 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:
time = 0
graph = defaultdict(list)
N = 0

def dfs(node):
nonlocal N

if not node: return

N += 1

if node.left:
graph[node.left.val].append(node.val)
graph[node.val].append(node.left.val)

if node.right:
graph[node.right.val].append(node.val)
graph[node.val].append(node.right.val)

dfs(node.left)
dfs(node.right)

dfs(root)

if N == 1: return 0

s = [start]
visited = defaultdict(lambda : False)
visited[start] = True

while s:
nxt = []

for node in s:
for nei in graph[node]:
if not visited[nei]:
visited[nei] = True
nxt.append(nei)

time += 1
s = nxt

return time - 1


• /**
* 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:
unordered_map<int, vector<int>> g;

int amountOfTime(TreeNode* root, int start) {
dfs(root);
queue<int> q{ {start} };
unordered_set<int> vis;
int ans = -1;
while (q.size()) {
++ans;
for (int n = q.size(); n; --n) {
int i = q.front();
q.pop();
vis.insert(i);
for (int j : g[i]) {
if (!vis.count(j)) {
q.push(j);
}
}
}
}
return ans;
}

void dfs(TreeNode* root) {
if (!root) return;
if (root->left) {
g[root->val].push_back(root->left->val);
g[root->left->val].push_back(root->val);
}
if (root->right) {
g[root->val].push_back(root->right->val);
g[root->right->val].push_back(root->val);
}
dfs(root->left);
dfs(root->right);
}
};

• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func amountOfTime(root *TreeNode, start int) int {
g := map[int][]int{}
var dfs func(*TreeNode)
dfs = func(root *TreeNode) {
if root == nil {
return
}
if root.Left != nil {
g[root.Val] = append(g[root.Val], root.Left.Val)
g[root.Left.Val] = append(g[root.Left.Val], root.Val)
}
if root.Right != nil {
g[root.Val] = append(g[root.Val], root.Right.Val)
g[root.Right.Val] = append(g[root.Right.Val], root.Val)
}
dfs(root.Left)
dfs(root.Right)
}

dfs(root)
q := []int{start}
ans := -1
vis := map[int]bool{}
for len(q) > 0 {
ans++
for n := len(q); n > 0; n-- {
i := q[0]
q = q[1:]
vis[i] = true
for _, j := range g[i] {
if !vis[j] {
q = append(q, j)
}
}
}
}
return ans
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function amountOfTime(root: TreeNode | null, start: number): number {
const map = new Map<number, number[]>();
const create = ({ val, left, right }: TreeNode) => {
if (left != null) {
map.set(val, [...(map.get(val) ?? []), left.val]);
map.set(left.val, [...(map.get(left.val) ?? []), val]);
create(left);
}
if (right != null) {
map.set(val, [...(map.get(val) ?? []), right.val]);
map.set(right.val, [...(map.get(right.val) ?? []), val]);
create(right);
}
};
create(root);
const dfs = (st: number, fa: number) => {
let res = 0;
for (const v of map.get(st) ?? []) {
if (v !== fa) {
res = Math.max(res, dfs(v, st) + 1);
}
}
return res;
};
return dfs(start, -1);
}



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).