Python greedy solution
To block x player, we only consider xnode\'s parent node or it\'s two children nodes as y node.\n\n    def btreeGameWinningMove(self, root: TreeNode, n: int, x:
To block x player, we only consider xnode\'s parent node or it\'s two children nodes as y node.\n```\n    def btreeGameWinningMove(self, root: TreeNode, n: int, x: int) -> bool:\n        def find(root, x):\n            if not root:\n                return\n            if root.val == x:\n                return root\n            \n            return find(root.left, x) or find(root.right, x)\n        \n        xnode = find(root, x)\n        \n        def count(node):\n            if not node:\n                return 0\n            return 1 + count(node.left) + count(node.right)\n        \n        l, r = count(xnode.left), count(xnode.right)\n        return n-l-r-1 > (l+r+1) or l > n-l or r > n - r\n```
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