Simple python O(n) solution that beats 100%
We need to count tree sizes of children of x node. \nThen the tree size of the rest of the tree could be calculated directly from these findings.\nx_parent = n 
We need to count **tree sizes** of children of x node. \nThen the tree **size of the rest of the tree** could be calculated directly from these findings.\nx_parent = n - x_left - x_right - 1\nIf any of the tree sizes gets the majority, it\'s possible for us to win:\ncapacity > n - capacity \n```\ndef btreeGameWinningMove(self, root, n, x):\n        tree = {\'x_left\' : 0, \'x_right\' : 0, \'x_parent\' : 0}\n        def dfs(node):\n            if not node:\n                return 0\n            left = dfs(node.left)\n            right = dfs(node.right)\n            if node.val == x:\n                tree[\'x_left\'] = left\n                tree[\'x_right\'] = right\n            return left + 1 + right\n        dfs(root)\n        \n        tree[\'x_parent\'] = n - tree[\'x_left\'] - 1 - tree[\'x_right\']\n        \n        for capacity in tree.values():\n            if capacity > n - capacity:\n                return True\n        return False\n```\nThis solution is O(n) time and O(n) space for recursion stack
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