Problem Description

You have a binary tree and need to install security cameras on some of its nodes. Each camera placed on a node can monitor:

• The node itself
• Its parent node (if it exists)
• Its immediate children nodes (if they exist)

Your task is to find the minimum number of cameras needed to monitor every single node in the tree.

For example, if you place a camera on a node, it will cover that node, the node directly above it, and the nodes directly below it. The goal is to strategically place cameras so that all nodes are monitored while using as few cameras as possible.

The solution uses dynamic programming with three states for each node:

• State a: The node has a camera installed on it
• State b: The node doesn't have a camera but is monitored by at least one of its children
• State c: The node doesn't have a camera and is not monitored by its children

The algorithm recursively calculates the minimum cameras needed for each state at every node, considering:

• If a node has a camera (a), both its children are automatically monitored, so we can choose any state for them
• If a node is monitored by its children (b), at least one child must have a camera
• If a node is not monitored by its children (c), both children must be monitored by their own children

The final answer is the minimum between placing a camera at the root or having the root monitored by its children.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: A binary tree is a special type of graph where each node has at most two children and there are no cycles.

Is it a tree?

• Yes: The problem explicitly states we're working with a binary tree structure.

DFS

• We arrive at DFS (Depth-First Search) as our algorithm choice.

Conclusion: The flowchart suggests using a DFS approach for solving the Binary Tree Cameras problem.

This makes perfect sense because:

1. We need to traverse the entire tree to determine camera placements
2. The solution requires us to make decisions at each node based on information from its children (bottom-up approach)
3. DFS allows us to recursively process subtrees and aggregate results back up to parent nodes
4. The dynamic programming states (whether a node has a camera, is monitored by children, or is not monitored) are naturally computed in a post-order DFS traversal where we process children before their parents

The DFS pattern enables us to:

• Visit each node exactly once
• Calculate the optimal camera placement for each subtree before deciding on parent nodes
• Propagate the monitoring states from leaves up to the root
• Choose the minimum number of cameras needed based on the three possible states at each node

Intuition

The key insight is to think about this problem from the bottom up - starting from the leaves and working our way to the root. Why? Because a camera's coverage extends both upward and downward, creating dependencies between parent and child nodes.

Consider what happens at each node. We have three possible situations:

1. Place a camera at this node
2. Don't place a camera, but get monitored by a child's camera
3. Don't place a camera and don't get monitored by children (rely on parent's camera)

The greedy approach of always placing cameras as high as possible doesn't work because we might miss optimal placements. Instead, we need to consider all possibilities and choose the best one.

Here's the critical observation: if we know the optimal camera placement for all subtrees, we can determine the optimal placement for the current node. This naturally leads to a recursive solution where we solve smaller subproblems first.

For each node, we need to track different states because the parent's decision depends on what happened with the children. If a child has a camera, it monitors its parent. If a child is unmonitored, the parent must either have a camera or be monitored by its other child.

The clever part is recognizing that we need exactly three states:

• State a: Having a camera here covers this node, its parent, and its children
• State b: Being monitored by children means at least one child must have a camera
• State c: Not being monitored by children means we're counting on the parent to monitor us

By calculating the minimum cameras needed for each state at every node and propagating these values up the tree, we can find the global minimum. The final answer is the minimum between placing a camera at the root (state a) or having the root monitored by its children (state b). We can't leave the root unmonitored (state c) since it has no parent to monitor it.

Learn more about Tree, Depth-First Search, Dynamic Programming and Binary Tree patterns.

Solution Approach

The implementation uses dynamic programming with DFS to solve this problem. Let's walk through how the solution works:

Function Structure: We define a recursive dfs(root) function that returns three values for each node:

• a: Minimum cameras needed if we place a camera at this node
• b: Minimum cameras needed if this node is monitored by its children
• c: Minimum cameras needed if this node is not monitored by its children

Base Case: When we reach a null node, we return [inf, 0, 0]. The inf for state a indicates it's impossible to place a camera on a null node. The zeros for states b and c mean no cameras are needed for non-existent nodes.

Recursive Case: For each non-null node, we first recursively solve for the left and right subtrees:

• la, lb, lc = dfs(root.left)
• ra, rb, rc = dfs(root.right)

Then we calculate the three states for the current node:

1. State a (camera at current node):

   • Formula: a = min(la, lb, lc) + min(ra, rb, rc) + 1
   • If we place a camera here, both children are automatically monitored
   • We can choose any state for the children (hence the min), plus 1 for the current camera

2. State b (monitored by children):

   • Formula: b = min(la + rb, lb + ra, la + ra)
   • At least one child must have a camera to monitor the parent
   • We consider three scenarios: left has camera, right has camera, or both have cameras
   • We pick the minimum among these options

3. State c (not monitored by children):

   • Formula: c = lb + rb
   • Neither child has a camera pointing at this node
   • Both children must be monitored by their own children (state b)

Final Answer: After computing the states for the root, we return min(a, b). We exclude state c because the root has no parent to monitor it, so it must either have a camera or be monitored by its children.

The algorithm runs in O(n) time since we visit each node once, and uses O(h) space for the recursion stack, where h is the height of the tree.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach. Consider this binary tree:

       1
      / \
     2   3
    /
   4

We'll compute the three states (a, b, c) for each node, working from the leaves up.

Step 1: Process leaf node 4

• Node 4 has no children, so dfs(null) returns [inf, 0, 0] for both left and right
• State a: min(inf, 0, 0) + min(inf, 0, 0) + 1 = 0 + 0 + 1 = 1 (camera at node 4)
• State b: min(inf + 0, 0 + inf, inf + inf) = min(inf, inf, inf) = inf (can't be monitored by non-existent children)
• State c: 0 + 0 = 0 (no children to monitor)
• Node 4 returns: [1, inf, 0]

Step 2: Process leaf node 3

• Node 3 has no children, similar to node 4
• Node 3 returns: [1, inf, 0]

Step 3: Process node 2

• Left child (node 4) returns [1, inf, 0], right child (null) returns [inf, 0, 0]
• State a: min(1, inf, 0) + min(inf, 0, 0) + 1 = 0 + 0 + 1 = 1 (camera at node 2)
• State b: min(1 + 0, inf + inf, 1 + inf) = min(1, inf, inf) = 1 (node 4 has camera, monitoring node 2)
• State c: inf + 0 = inf (node 4 must be in state b, which is impossible)
• Node 2 returns: [1, 1, inf]

Step 4: Process root node 1

• Left child (node 2) returns [1, 1, inf], right child (node 3) returns [1, inf, 0]
• State a: min(1, 1, inf) + min(1, inf, 0) + 1 = 1 + 0 + 1 = 2 (camera at node 1, optimal states for children)
• State b: min(1 + inf, 1 + 1, 1 + 1) = min(inf, 2, 2) = 2 (either child has camera)
• State c: 1 + inf = inf (impossible since node 3 can't be in state b)
• Node 1 returns: [2, 2, inf]

Final Answer: min(a, b) = min(2, 2) = 2

The optimal solution uses 2 cameras. One valid placement is:

• Camera at node 2 (monitors nodes 1, 2, and 4)
• Camera at node 3 (monitors node 3)

This example shows how the algorithm considers all possibilities at each node and propagates the optimal solutions upward, ultimately finding the minimum number of cameras needed.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The algorithm performs a depth-first search (DFS) traversal of the binary tree. Each node is visited exactly once during the traversal. At each node, the function performs constant-time operations:

• Recursive calls to left and right children
• Computing values a, b, and c using simple arithmetic operations and min() comparisons

Since we visit all n nodes exactly once and perform O(1) work at each node, the overall time complexity is O(n).

Space Complexity: O(n)

The space complexity is dominated by the recursive call stack. In the worst case, when the binary tree is completely unbalanced (forming a linked list), the recursion depth can reach n levels, requiring O(n) space on the call stack. In the best case of a perfectly balanced tree, the recursion depth would be O(log n), but we consider the worst-case scenario for space complexity analysis.

Additionally, the function uses only a constant amount of extra space for variables (la, lb, lc, ra, rb, rc, a, b, c) at each recursive level, which doesn't change the overall space complexity.

Therefore, the space complexity is O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrectly Handling the Root Node's Final State

Pitfall: A common mistake is returning min(a, b, c) or just a for the final answer, forgetting that the root node has no parent to monitor it.

Why it's wrong: State c represents a node that is NOT monitored. Since the root has no parent, if we choose state c, the root would remain unmonitored forever, violating the problem's requirement.

Solution: Always return min(a, b) for the root, excluding state c:

camera_at_root, root_covered, root_not_covered = dfs(root)
# Correct: Root must be monitored
return min(camera_at_root, root_covered)
# Wrong: Would leave root potentially unmonitored
# return min(camera_at_root, root_covered, root_not_covered)

2. Incorrect Base Case for Null Nodes

Pitfall: Returning incorrect values for null nodes, such as [0, 0, 0] or [1, 0, 0].

Why it's wrong: The first value must be inf because you cannot place a camera on a non-existent node. Using 0 or 1 would incorrectly suggest it's possible or cheap to place a camera there.

Solution: Always return [inf, 0, 0] for null nodes:

if node is None:
    return inf, 0, 0  # Correct
    # return 0, 0, 0  # Wrong: suggests we can place a camera on null

3. Misunderstanding State b Calculation

Pitfall: Calculating state b as simply la + ra or min(la, ra), forgetting that at least one child MUST have a camera to monitor the parent.

Why it's wrong: State b means the node is monitored by its children. This requires at least one child to have a camera pointing upward.

Solution: Consider all valid combinations where at least one child has a camera:

# Correct: At least one child must have a camera
covered_no_camera = min(
    left_camera + right_covered,   # left has camera
    left_covered + right_camera,   # right has camera  
    left_camera + right_camera     # both have cameras
)
# Wrong: Doesn't ensure a child has a camera
# covered_no_camera = left_covered + right_covered

4. Forgetting to Add 1 When Placing a Camera

Pitfall: When calculating state a, forgetting to add 1 for the camera being placed at the current node.

Why it's wrong: Each camera costs 1, so we must account for it in our count.

Solution: Always add 1 when placing a camera:

# Correct
camera_here = min(la, lb, lc) + min(ra, rb, rc) + 1
# Wrong: Forgets to count the current camera
# camera_here = min(la, lb, lc) + min(ra, rb, rc)

5. Using Global Variables Incorrectly

Pitfall: Some might try to use a global counter or mutable default arguments to track cameras, leading to incorrect results when the function is called multiple times.

Solution: Use the return value approach with tuples to maintain state properly:

# Correct: Return states as tuple
def dfs(node):
    # ... calculations ...
    return camera_here, covered_no_camera, not_covered

# Wrong: Using global counter
# self.cameras = 0  # This persists between test cases!