LeetCode #2385 — MEDIUM

Amount of Time for Binary Tree to Be Infected

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Tree

Example 1

[1,5,3,null,4,10,6,9,2]
3

Example 2

[1]
1

Core Insight

What unlocks the optimal approach

Convert the tree to an undirected graph to make it easier to handle.
Use BFS starting at the start node to find the distance between each node and the start node. The answer is the maximum distance.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2385: Amount of Time for Binary Tree to Be Infected
/**
 * 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;
    }
}

// Accepted solution for LeetCode #2385: Amount of Time for Binary Tree to Be Infected
/**
 * 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, *TreeNode)
	dfs = func(node, fa *TreeNode) {
		if node == nil {
			return
		}
		if fa != 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)
	}
	var dfs2 func(int, int) int
	dfs2 = func(node, fa int) (ans int) {
		for _, nxt := range g[node] {
			if nxt != fa {
				ans = max(ans, 1+dfs2(nxt, node))
			}
		}
		return
	}
	dfs(root, nil)
	return dfs2(start, -1)
}

# Accepted solution for LeetCode #2385: 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:
        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)

// Accepted solution for LeetCode #2385: Amount of Time for Binary Tree to Be Infected
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2385: Amount of Time for Binary Tree to Be Infected
// /**
//  * 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;
//     }
// }

// Accepted solution for LeetCode #2385: Amount of Time for Binary Tree to Be Infected
/**
 * 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 g: Map<number, number[]> = new Map();
    const dfs = (node: TreeNode | null, fa: TreeNode | null) => {
        if (!node) {
            return;
        }
        if (fa) {
            if (!g.has(node.val)) {
                g.set(node.val, []);
            }
            g.get(node.val)!.push(fa.val);
            if (!g.has(fa.val)) {
                g.set(fa.val, []);
            }
            g.get(fa.val)!.push(node.val);
        }
        dfs(node.left, node);
        dfs(node.right, node);
    };
    const dfs2 = (node: number, fa: number): number => {
        let ans = 0;
        for (const nxt of g.get(node) || []) {
            if (nxt !== fa) {
                ans = Math.max(ans, 1 + dfs2(nxt, node));
            }
        }
        return ans;
    };
    dfs(root, null);
    return dfs2(start, -1);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.