What is the time complexity of Binary Tree Cameras?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Binary Tree Cameras?

This problem uses Dynamic Programming, Tree, Depth-First Search, Binary Tree. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Binary Tree Cameras?

Insight 1: Use Dynamic Programming and Depth-First Search (DFS). DFS allows us to traverse the tree, and Dynamic Programming helps us make optimal camera placement decisions at each node by considering the states of its children. Insight 2: Define three states for each node: 0 (not covered), 1 (has camera), and 2 (covered by a camera on a child or parent). This simplifies the logic for making camera placement decisions.

Which companies ask LeetCode 968. Binary Tree Cameras in interviews?

This problem is frequently asked at DE Shaw, DP world, Graviton, MathWorks, PhonePe and 3 more companies.

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Binary Tree Cameras - LeetCode 968 Solution

Binary Tree Cameras - Complete Solution Guide

Binary Tree Cameras is LeetCode problem 968, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given the root of a binary tree. We install cameras on the tree nodes where each camera at a node can monitor its parent, itself, and its immediate children. Return the minimum number of cameras needed to monitor all nodes of the tree . Example 1: Input: root = [0,0,null,0,0] Output: 1 Explanation: One camera is enough to monitor all nodes if placed as shown. Example 2: Input: root = [0,0,null,0,null,0,null,null,0] Output: 2 Explanation: At least two cameras are needed to monitor all nod

Detailed Explanation

The problem asks us to find the minimum number of cameras needed to monitor all nodes in a binary tree. Each camera placed at a node can monitor its parent, itself, and its children. Essentially, every node needs to be 'covered' either by a camera directly placed on it, on its parent, or on one of its children. The input is the root of the binary tree, and the output is the minimum number of cameras required.

Solution Approach

The provided solution uses a DFS traversal of the binary tree. During the traversal, for each node, we determine its state and the number of cameras required in its subtree based on the states of its children. The `dfs` function returns a tuple (or array in Java and struct in C/C++) representing the state of the current node and the number of cameras in its subtree. We use three states: 0 (node is not covered), 1 (node has a camera), and 2 (node is covered by a camera on a child or parent). By analyzing the states of the left and right children, the algorithm strategically places cameras to minimize their overall count. The crucial part is realizing that not every node *needs* a camera; if it's covered by one of its children, we can save a camera. The bottom-up nature ensures optimal decisions are made.

Step-by-Step Algorithm

Step 1: Define the DFS function that takes a node as input and returns a tuple (or similar structure) representing the node's state and the number of cameras in its subtree.
Step 2: Handle the base case: if the node is null, return (2, 0), indicating that it is covered and requires no cameras.
Step 3: Recursively call the DFS function on the left and right children to obtain their states and camera counts.
Step 4: Calculate the number of cameras needed based on the children's states. If either child is not covered (state 0), place a camera on the current node (increment camera count and set current node's state to 1).
Step 5: If the current node does not have a camera, check if either child has a camera. If so, the current node is covered (set current node's state to 2).
Step 6: If neither child has a camera and the current node does not have a camera, the current node is not covered (set current node's state to 0).
Step 7: After the DFS traversal, check the state of the root node. If the root node is not covered (state 0), place a camera on it (increment camera count).
Step 8: Return the final camera count.

Key Insights

Insight 1: Use Dynamic Programming and Depth-First Search (DFS). DFS allows us to traverse the tree, and Dynamic Programming helps us make optimal camera placement decisions at each node by considering the states of its children.
Insight 2: Define three states for each node: 0 (not covered), 1 (has camera), and 2 (covered by a camera on a child or parent). This simplifies the logic for making camera placement decisions.
Insight 3: A bottom-up approach during DFS is crucial. We determine the camera placement for the leaves before considering their parents. The base case of an empty (null) node is considered as being covered (state 2) because it doesn't require a camera.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Dynamic Programming, Tree, Depth-First Search, Binary Tree.

Companies

Asked at: DE Shaw, DP world, Graviton, MathWorks, PhonePe, Quora, Visa, eBay.

Binary Tree Cameras | LeetCode 968

Hard

Dynamic Programming Tree Depth-First Search Binary Tree

Time: O(n)Space: O(n)

Solution Code

Available in:

/**
 * 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 {
    public int minCameraCover(TreeNode root) {
        int[] result = dfs(root);
        return result[0] + (result[1] == 0 ? 1 : 0);
    }

    private int[] dfs(TreeNode node) {
        // Returns an array: [number of cameras in subtree, state of node]
        // States:
        // 0: Node is not monitored, needs a camera from its parent.
        // 1: Node has a camera.
        // 2: Node is monitored by another camera.

        if (node == null) {
            return new int[]{0, 2};
        }

        int[] leftResult = dfs(node.left);
        int[] rightResult = dfs(node.right);

        int leftCameras = leftResult[0];
        int leftState = leftResult[1];
        int rightCameras = rightResult[0];
        int rightState = rightResult[1];

        int cameras = leftCameras + rightCameras;

        if (leftState == 0 || rightState == 0) {
            cameras += 1;
            return new int[]{cameras, 1};
        }

        if (leftState == 1 || rightState == 1) {
            return new int[]{cameras, 2};
        }

        return new int[]{cameras, 0};
    }
}

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Problem Info

Description

Problem 968: Binary Tree Cameras - Detailed explanation and solution approaches.

Topics

Dynamic Programming Tree Depth-First Search Binary Tree

Companies

DE ShawDP worldGravitonMathWorksPhonePeQuoraVisaeBay

Stats

Difficulty:Hard

Languages:4

Companies:8

Problem #:968

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Define the DFS function that takes a node as input and returns a tuple (or similar structure) representing the node's state and the number of cameras in its subtree.
2
Step 2: Handle the base case: if the node is null, return (2, 0), indicating that it is covered and requires no cameras.
3
Step 3: Recursively call the DFS function on the left and right children to obtain their states and camera counts.
4
Step 4: Calculate the number of cameras needed based on the children's states. If either child is not covered (state 0), place a camera on the current node (increment camera count and set current node's state to 1).
5
Step 5: If the current node does not have a camera, check if either child has a camera. If so, the current node is covered (set current node's state to 2).
6
Step 6: If neither child has a camera and the current node does not have a camera, the current node is not covered (set current node's state to 0).
7
Step 7: After the DFS traversal, check the state of the root node. If the root node is not covered (state 0), place a camera on it (increment camera count).
8
Step 8: Return the final camera count.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because we visit each node of the tree exactly once during the DFS traversal. The space complexity is determined by the maximum depth of the recursion stack. In the worst case, where the tree is a linked list, the recursion can go as deep as n, resulting in O(n) space complexity. In a balanced tree, the space complexity would be O(log n).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Greedy Approach

Attempting to solve this problem with a purely greedy approach (e.g., always placing a camera on a node if its children aren't covered) might seem intuitive but will likely fail. A greedy approach doesn't consider the global optimal solution, leading to more cameras than necessary in some cases. It is difficult to prove any greedy approach is optimal.

Time: O(n)Space: O(n) in worst case for recursive implementation.

Trade-offs: While simpler to conceptualize initially, a greedy approach is unlikely to produce the optimal solution consistently. The dynamic programming with DFS approach guarantees the minimum number of cameras.

✨

Key Insights

1Insight 1: Use Dynamic Programming and Depth-First Search (DFS). DFS allows us to traverse the tree, and Dynamic Programming helps us make optimal camera placement decisions at each node by considering the states of its children.
2Insight 2: Define three states for each node: 0 (not covered), 1 (has camera), and 2 (covered by a camera on a child or parent). This simplifies the logic for making camera placement decisions.
3Insight 3: A bottom-up approach during DFS is crucial. We determine the camera placement for the leaves before considering their parents. The base case of an empty (null) node is considered as being covered (state 2) because it doesn't require a camera.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty tree (root is null). The code handles this correctly by returning (2, 0), indicating that an empty tree is considered covered and requires no cameras.
Edge case 2: Single node tree. If the root has no children, the DFS will return a state of 0 for the root since both left and right are considered covered(2). The final check `if root_state == 0` will add a camera to the root, which is the correct answer.
Edge case 3: Skewed tree (all nodes on one side). The algorithm still works correctly because the bottom-up DFS approach ensures that cameras are placed strategically regardless of the tree's shape.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling the base case of null nodes. Assuming that null nodes are not covered can lead to unnecessary camera placements.

✗Common mistake 2: Implementing the DFS without a bottom-up approach. Making camera placement decisions at higher levels of the tree before considering the lower levels can result in suboptimal solutions.

✗Common mistake 3: Not considering all three states (not covered, has camera, covered). Failing to distinguish between these states can lead to flawed logic in determining camera placement.

✗Common mistake 4: Forgetting the final check for the root node's state after the DFS traversal. The root node might still need a camera if it's not covered by its children.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Surveillance system design. Imagine you need to install security cameras to monitor a network of interconnected areas. The tree structure can represent dependencies and visibility ranges, and the algorithm helps minimize the number of cameras while ensuring complete coverage.

🏦Real-world scenario 2: Network monitoring. Consider a network of computers or sensors arranged in a hierarchical structure. The algorithm can be used to determine the minimum number of monitoring devices needed to ensure that all parts of the network are observed.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 8, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I choose between recursive and iterative tree traversal?

Use recursive traversal for cleaner code when stack depth isn't a concern (tree height < ~1000). Use iterative with explicit stack for very deep trees to avoid stack overflow. Iterative is also preferred when you need fine-grained control over the traversal process.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Binary Tree Cameras"?

For this hard-level Dynamic Programming problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether tree techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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