How do you solve Binary Tree Cameras on LeetCode?

Binary Tree Cameras (LeetCode #968) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(h) space complexity. Key insight: Cameras can cover the node itself, its parent, and its children.

What is the brute force solution for Binary Tree Cameras?

The brute force approach for Binary Tree Cameras is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible placement of cameras in the tree. This means we would try to place a camera at each node and see if it covers all other nodes, which can be very inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Binary Tree Cameras?

Binary Tree Cameras uses the following concepts: Dynamic Programming, Tree, Depth-First Search, Binary Tree. The recommended patterns to study are: Depth-First Search, Dynamic Programming.

What are the interview tips for Binary Tree Cameras?

Always clarify the problem requirements before diving into coding. Think about edge cases, such as a single-node tree. Explain your thought process while coding to demonstrate your understanding.

What are common mistakes when solving Binary Tree Cameras?

Not considering all states of a node (covered, needs camera, has camera). Failing to account for the root node's coverage.

#968

Binary Tree Cameras

Hard

Dynamic ProgrammingTreeDepth-First SearchBinary TreeDepth-First SearchDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(h)

💡

Intuition

Time O(n)Space O(h)

The optimal solution uses a depth-first search (DFS) approach with a state-based system to determine if a node is covered, needs a camera, or has a camera. This allows us to efficiently calculate the minimum number of cameras needed.

⚙️

Algorithm

3 steps

1Step 1: Perform a DFS traversal of the tree.
2Step 2: For each node, determine its state: covered, needs a camera, or has a camera.
3Step 3: Count the cameras based on the states of the child nodes.

solution.py23 lines

1# Full working Python code
2class TreeNode:
3    def __init__(self, val=0, left=None, right=None):
4        self.val = val
5        self.left = left
6        self.right = right
7
8class Solution:
9    def minCameraCover(self, root: TreeNode) -> int:
10        self.cameras = 0
11        def dfs(node):
12            if not node:
13                return 1
14            left = dfs(node.left)
15            right = dfs(node.right)
16            if left == 0 or right == 0:
17                self.cameras += 1
18                return 2
19            return 1 if left == 2 or right == 2 else 0
20        if dfs(root) == 0:
21            self.cameras += 1
22        return self.cameras
23

ℹ

Complexity note: The time complexity is O(n) because we visit each node once. The space complexity is O(h) due to the recursion stack, where h is the height of the tree.

1Cameras can cover the node itself, its parent, and its children.
2A node needs a camera if any of its children are not covered.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.