VERY EASY 0 MS SOLUTION
\n/**\n * Definition for a binary tree node.\n * public class TreeNode {\n *     int val;\n *     TreeNode left;\n *     TreeNode right;\n *     TreeNode() {}\n
```\n/**\n * Definition for a binary tree node.\n * public class TreeNode {\n *     int val;\n *     TreeNode left;\n *     TreeNode right;\n *     TreeNode() {}\n *     TreeNode(int val) { this.val = val; }\n *     TreeNode(int val, TreeNode left, TreeNode right) {\n *         this.val = val;\n *         this.left = left;\n *         this.right = right;\n *     }\n * }\n */\nclass Solution {\n        int left=0;\n        int right=0;\n    public boolean btreeGameWinningMove(TreeNode root, int n, int x) {\n   int total=     dfs(root,x);\n            \n     int parent=  total-left-right-1;\n            \n\n            if(parent>left+right){\n                    return true;\n            }else if(left>parent+right){\n                    return true;\n            }else if(right>parent+left){\n                    return true;\n            }\n            \n            return false;\n            \n    }\n        public int dfs(TreeNode root,int x){\n                \n                if(root==null)return 0;\n                \n              int l=  dfs(root.left,x);\n              int r=  dfs(root.right,x);\n                \n                \n                if(root.val==x){\n                left=l;\n                right=r;\n                }\n                \n                \n        return l+r+1;\n                \n                \n        }\n\n        \n}\n\n\n```
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