Python based on left, right and above
The first step is to calculate the size of the subtree rooted at each node.\nOnce we have this info, it boils down to we as Player 2 choosing the node, that is 
The first step is to calculate the size of the subtree rooted at each node.\nOnce we have this info, it boils down to we as Player 2 choosing the node, that is either parent of X, or left child of X  or right child of X.\n\nLets definte "Above" = number of nodes that are not in my subtree.\n\nIf we choose the parent of X =. we can color everythin Above, while X can color his left and right subtree.\nIf we choose X\'s left subtree, then X can choose his right subtree + Above\nIf we choose X\'s right subtree, then X can shoose his left subtree + Above.\n\nFinally  we see if we can win, by choosing the best of the above 3 options.\n\n```\ndef btreeGameWinningMove(self, root: TreeNode, n: int, x: int) -> bool:\n        def traverse(node):\n            if not node: return 0\n            res = 1 + traverse(node.left) + traverse(node.right)\n            d[node.val] = res\n            return res\n        \n        def find(node, above):\n            if not node: return False\n            leftval = node.left.val if node.left else 0\n            rightval = node.right.val if node.right else 0\n            if node.val == x:\n                return above > d[node.val] or d[rightval] > above + d[leftval] or d[leftval] > above + d[rightval]\n            return find(node.left, 1 + above + d[rightval]) or find(node.right, 1 + above + d[leftval])\n        \n        d = defaultdict(int)\n        traverse(root)\n        return find(root, 0)\n```
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