Binary Tree Cameras — LeetCode 968 Python Solution

The problem

You are given the root of a binary tree. We install cameras on the tree nodes where each camera at a node can monitor its parent, itself, and its immediate children.

Example

Input

root = [0,0,null,0,0]

Output

1

Explanation

One camera is enough to monitor all nodes if placed as shown.

Python solution

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def minCameraCover(self, root: Optional[TreeNode]) -> int:
        def dfs(root):
            if root is None:
                return inf, 0, 0
            la, lb, lc = dfs(root.left)
            ra, rb, rc = dfs(root.right)
            a = min(la, lb, lc) + min(ra, rb, rc) + 1
            b = min(la + rb, lb + ra, la + ra)
            c = lb + rb
            return a, b, c

        a, b, _ = dfs(root)
        return min(a, b)

Complexity

Pattern: Tree Traversal

Choose the order — preorder, inorder, postorder, level — and the problem solves itself. LeetCode 968. Binary Tree Cameras is filed here because LeetCode tags it Tree and Binary Tree, which is the vocabulary this hub collects.

The tree traversal guide has the Python template for the pattern and the 225 LeetCode problems that use it.

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