Java Solution
\n\t// value of left node of x.\n    int left = 0;\n    // value of right node of x.\n    int right = 0;\n    public boolean btreeGameWinningMove(TreeNode root,
```\n\t// value of left node of x.\n    int left = 0;\n    // value of right node of x.\n    int right = 0;\n    public boolean btreeGameWinningMove(TreeNode root, int n, int x) {\n        if (root.left == null && root.right == null) {\n            return false;\n        }\n        int[] numOfChildren = new int[n];\n        traverse(root, numOfChildren, x);\n        if ((n - (numOfChildren[x - 1])) > (numOfChildren[x - 1])) {\n\t\t\t// We can choose the nodes that do not belong to subtree of x.\n            return true;\n        } else {\n\t\t\t// We can choose one of the subtrees of x.\n            if (left != 0 && numOfChildren[left - 1] > n - numOfChildren[left - 1]) {\n                return true;\n            }\n            if (right != 0 && numOfChildren[right - 1] > n - numOfChildren[right - 1]) {\n                return true;\n            }\n        }\n        return false;\n    }\n\n    // Find the number of nodes for each subtree.\n    private int traverse(TreeNode root, int[] numOfChildren, int x) {\n        if (root == null) {\n            return 0;\n        }\n        if (root.val == x) {\n            if (root.left != null) {\n                left = root.left.val;\n            }\n            if (root.right != null) {\n                right = root.right.val;\n            }\n        }\n        int left = traverse(root.left, numOfChildren, x);\n        int right = traverse(root.right, numOfChildren, x);\n        numOfChildren[root.val - 1] = left + right + 1;\n        return left + right + 1;\n    }\n```
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