Can the Second Tree-Coloring Player Force a Majority?

### Problem A binary tree has n distinct values 1 through n, with n odd. The first player initially colors node x red. You choose another node y to color blue. Starting with red, players alternate coloring an uncolored parent or child of any node already their own color. A player passes only if no such move exists; play ends when both pass. Return whether some legal y guarantees blue more nodes than red. ```cpp class Solution { public: bool btreeGameWinningMove(TreeNode* root, int n, int x); }; ``` ### Examples **Example 1:** ```text Input: root = [2,1,3], n = 3, x = 1 Output: true Explanation: Choose node2. Blue then controls the two-node side containing2 and3, beyond red node1. ``` **Example 2:** ```text Input: root = [2,1,3], n = 3, x = 2 Output: false Explanation: Either legal blue seed lies in a one-node component, while red can own the other two nodes. ``` ### Constraints - The tree has exactly n nodes, with `1 <= x <= n <= 100`. - n is odd. - Node values are distinct integers from 1 through n. ### Contract considerations Retain the native TreeNode interface. Return whether a winning choice exists, not a replacement graph or seed transcript. ```hint Read the boundary condition When n is one, is there any legal second starting node distinct from x? ``` ### Boundary reminders Retain the native TreeNode interface. Return whether a winning choice exists, not a replacement graph or seed transcript.