House robber III

We will traverse the tree in DFS and take both with and without root result. Then we calculate the max from both of them and return the result.

# 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 rob(self, root: Optional[TreeNode]) -> int:
        def dfs(root):
            if not root:
                return (0, 0)

            left = dfs(root.left)
            right = dfs(root.right)

            with_root = root.val + left[1] + right[1]
            without_root = max(left) + max(right)

            return (with_root, without_root)

        return max(dfs(root))

Time Complexity: O(n)
Space Complexity: O(n)