Simple Python DFS Solution
Intuition\nThe problem involves finding and counting the number of subtrees in a binary tree where the average value of the subtree is equal to the value of the
# Intuition\nThe problem involves finding and counting the number of subtrees in a binary tree where the average value of the subtree is equal to the value of the root of that subtree.\n\n# Approach\nThe approach taken here is a recursive one. The function `helper` is a recursive function that takes a node as input and returns a tuple `(count, total)`. The `count` represents the number of nodes in the subtree rooted at the given node, and `total` represents the sum of values of all nodes in the subtree.\n\nThe main function initializes a counter `self.count` to 0 and then calls the `helper` function on the root of the tree. The `helper` function is responsible for traversing the tree recursively and updating the count whenever it finds a subtree with the desired property.\n\nThe code efficiently counts the number of subtrees with the specified property by recursively calculating the count and total for each subtree in a bottom-up manner. The use of a tuple `(count, total)` allows the function to propagate necessary information up the recursive call stack efficiently.\n\n# Complexity\n- Time complexity: $$O(n)$$, where n is the number of nodes in the tree. Each node is visited once.\n- Space complexity: $$O(h)$$, where h is the height of the tree. The space complexity is determined by the depth of the recursion stack. In the worst case, for a skewed tree, it could be $$O(n)$$. In a balanced tree, it would be $$O(\\log n)$$.\n\n# Code\n```python\nclass Solution(object):\n    def averageOfSubtree(self, root):\n        self.count = 0\n        \n        def helper(node):\n            if not node:\n                return (0, 0)\n            \n            left_count, left_total = helper(node.left)\n            right_count, right_total = helper(node.right)\n            \n            count = 1 + left_count + right_count\n            total = node.val + left_total + right_total\n            \n            if node.val == total // count:\n                self.count += 1\n            \n            return (count, total)\n        \n        helper(root)\n        return self.count\n```
['Tree', 'Depth-First Search', 'Recursion', 'Binary Tree', 'Python']