Python 3   ||   12 lines,  w/ explanation and example   ||   T/S: 98% / 93%
The problem reduces to whether any of the three subgraphs with edges to node x have at least (n+1)//2 nodes.\n\nHere\'s the plan:\n- Traverse the tree with dfs 
The problem reduces to whether any of the three subgraphs with edges to node `x` have at least `(n+1)//2` nodes.\n\nHere\'s the plan:\n- Traverse the tree with `dfs` recursively.\n- For each `node`, rewrite `node.val` with the count of nodes in its subtree.\n- Evaluate whether `node` is x, and if so, determine whether one of its neighbors is greater than `(n+1)//2`.\n- Return `True` if so, of if either child returns `True`. If not, return `False`\n```\nclass Solution:\n    def btreeGameWinningMove(self, root: TreeNode, n: int, x: int) -> bool:\n\n        most = (n+1)//2\n\n        def dfs(node):\n            if not node: return False\n\n            if dfs(node.left) or dfs(node.right): return True\n            is_x = node.val == x\n            \n            l = node.left .val if node.left  else 0\n            r = node.right.val if node.right else 0\n\n            if is_x and (l >= most or r >= most): return True\n\n            node.val = 1 + l + r\n\n            if is_x: return node.val < most\n\n            return False\n\n        return dfs(root)\n```\n[https://leetcode.com/problems/binary-tree-coloring-game/submissions/1274328962/](https://leetcode.com/problems/binary-tree-coloring-game/submissions/1274328962/)\n\nI could be wrong, but I think that time complexity is *O*(*N*) and space complexity is *O*(*N*), in which *N* ~ the number of nodes.
['Python3']