The "Binary Tree Cameras interview question" is a challenging "Hard" problem. You need to place the minimum number of cameras on the nodes of a binary tree such that every node in the tree is monitored. A camera at a node monitors itself, its parent, and its immediate children. This is a variation of the "Vertex Cover" problem but specifically for trees.
Tech giants like Google, Adobe, and Meta use the "Binary Tree Cameras coding problem" to evaluate a candidate's ability to use Greedy recursion or Dynamic Programming on Trees. It requires you to think about "local" decisions that lead to a "global" optimum. It tests whether you can identify the three possible states of a node (has camera, is covered, needs coverage).
This problem is best solved using Greedy Depth-First Search. The strategy is to place cameras as high up as possible. We start from the leaves and move up.
0: Node is not covered (needs a camera from its parent).1: Node has a camera.2: Node is covered (but doesn't have a camera).0 (uncovered), the current node must have a camera. Increment camera count and return 1.1 (has camera), the current node is now covered. Return 2.0.0, it needs its own camera.Tree: A root with two leaf children.
0 (I'm a leaf, I'm not covered).0 (I'm a leaf, I'm not covered).1.
Total cameras: 1. (A camera at the root covers the root and both leaves).For tree optimization problems, always consider a "bottom-up" approach using post-order DFS. Leaves are the most constrained part of the tree, so making decisions based on them often leads to the global minimum. This is a core "Tree interview pattern" for "Hard" problems.
| Title | Difficulty | Topics | LeetCode |
|---|---|---|---|
| Binary Tree Maximum Path Sum | Hard | Solve | |
| Minimum Flips in Binary Tree to Get Result | Hard | Solve | |
| Longest ZigZag Path in a Binary Tree | Medium | Solve | |
| House Robber III | Medium | Solve | |
| Maximum Sum BST in Binary Tree | Hard | Solve |