The O(n) Club: Binary Tree Cameras: Minimum Coverage, Maximum Laziness

The O(n) Club: Binary Tree Cameras—Minimum Coverage, Maximum Laziness

⚡ TL;DR

How many cameras do you need to keep an eye on every spot in your binary tree? (Answer: fewer than most devs think.) Brute force is an invitation to regret. Greedy DFS takes the crown—here’s the Java sneak peek:

int minCameraCover(TreeNode root) {
    cameraCount = 0;
    return (dfs(root) == 2 ? ++cameraCount : cameraCount);
}
// 0 = Covered, 1 = Has Camera, 2 = Needs Camera
int dfs(TreeNode node) {
    if (node == null) return 0;
    int left = dfs(node.left), right = dfs(node.right);
    if (left == 2 || right == 2) {
        cameraCount++;
        return 1;
    }
    return (left == 1 || right == 1) ? 0 : 2;
}

🧠 How Devs Usually Mess This Up

So many folks pepper cameras onto every leaf, thinking more hardware equals more safety. That’s called ‘maximum budget, minimal thinking.’ Or they panic over node values, which matter exactly zero percent here. And when someone forgets those null children are always ‘covered’? Welcome to bug city.

🔍 What’s Actually Going On

Don’t toss a camera in every broom closet. Think like a security chief: each camera watches its node, its parent, and all its kids. Recursively, you let coverage bubble up from the leaves. Only add cameras where coverage gaps would otherwise start. With three states (has camera, covered, needs camera), you can play defense like a lazy chess master.

• Has Camera: This node carries the gear.
• Covered: Under surveillance, not responsible for anyone.
• Needs Camera: Yelling for help.

🛠️ PseudoCode

• Assign each node a state: HAS_CAMERA, COVERED, or NEEDS_CAMERA.
• Use post-order DFS to handle children first, then the current node.
• If any child needs a camera, install one here.
• If a child has a camera, be happy—you’re covered.
• If both kids are just covered, you might be the new problem—return NEEDS_CAMERA.
• null children are always covered (just like the QA server, apparently).

// Pseudo-logic
if (node == null):
    return COVERED
left = dfs(node.left)
right = dfs(node.right)
if (left == NEEDS_CAMERA or right == NEEDS_CAMERA):
    cameraCount++
    return HAS_CAMERA
if (left == HAS_CAMERA or right == HAS_CAMERA):
    return COVERED
return NEEDS_CAMERA

💻 The Code

class Solution {
    private int cameraCount = 0;
    // 0 = Covered, 1 = Has Camera, 2 = Needs Camera
    private int dfs(TreeNode node) {
        if (node == null) return 0;
        int left = dfs(node.left);
        int right = dfs(node.right);
        if (left == 2 || right == 2) {
            cameraCount++;
            return 1;
        }
        if (left == 1 || right == 1) {
            return 0;
        }
        return 2;
    }
    public int minCameraCover(TreeNode root) {
        cameraCount = 0;
        return dfs(root) == 2 ? cameraCount + 1 : cameraCount;
    }
}
 class TreeNode {
    int val;
    TreeNode left, right;
    TreeNode(int x) { val = x; }
}

⚠️ Pitfalls, Traps & Runtime Smacks

• Don’t camera every leaf: Unless you love wasting money or writing redundant code.
• Ignore node values: Structural problem, not a value contest.
• Handle null children: Null means already safe—chill out.
• Multiple placements can be optimal: Don’t sweat the arrangement, focus on the count.
• Performance: O(n) time, O(h) space for the recursion stack (h = tree height, so avoid palm trees).

🧠 Interviewer Brain-Tattoo

“When your instinct wants to tape a camera to every node, remember: putting cameras up high watches more—and that includes you during your next code review.”