LeetCode 968: Binary Tree Cameras Solution

Master LeetCode problem 968 (Binary Tree Cameras), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

Problem Explanation

Explanation:

To solve this problem, we can perform a depth-first search (DFS) traversal of the binary tree. We can keep track of three states for each node:

  • 0: The node is not covered by any camera.
  • 1: The node is covered by a camera.
  • 2: The node has a camera installed.

We can define a recursive function to traverse the tree and return the minimum number of cameras required to cover all nodes. At each node, we have the following cases:

  • If the node is a leaf node, we don't need to install a camera at this node, so we return 0.
  • If the node's children are covered and the node itself is not covered, we install a camera at this node and return 2.
  • If the node has a child that needs a camera, we install a camera at this node and return 2.
  • If the node is covered by a camera, we return 1.

We can start the DFS traversal from the root node and return the minimum of the number of cameras required for the root node being covered or not covered.

Time complexity: O(N), where N is the number of nodes in the tree. Space complexity: O(H), where H is the height of the tree.

:

Solution Code

class Solution {
    private int cameras = 0;
    
    public int minCameraCover(TreeNode root) {
        if (dfs(root) == 0) {
            cameras++;
        }
        return cameras;
    }
    
    private int dfs(TreeNode node) {
        if (node == null) {
            return 1;
        }
        
        int left = dfs(node.left);
        int right = dfs(node.right);
        
        if (left == 0 || right == 0) {
            cameras++;
            return 2;
        } else if (left == 2 || right == 2) {
            return 1;
        } else {
            return 0;
        }
    }
}

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