<p>Two players play a turn based game on a binary tree. We are given the <code>root</code> of this binary tree, and the number of nodes <code>n</code> in the tree. <code>n</code> is odd, and each node has a distinct value from <code>1</code> to <code>n</code>.</p>

<p>Initially, the first player names a value <code>x</code> with <code>1 &lt;= x &lt;= n</code>, and the second player names a value <code>y</code> with <code>1 &lt;= y &lt;= n</code> and <code>y != x</code>. The first player colors the node with value <code>x</code> red, and the second player colors the node with value <code>y</code> blue.</p>

<p>Then, the players take turns starting with the first player. In each turn, that player chooses a node of their color (red if player 1, blue if player 2) and colors an <strong>uncolored</strong> neighbor of the chosen node (either the left child, right child, or parent of the chosen node.)</p>

<p>If (and only if) a player cannot choose such a node in this way, they must pass their turn. If both players pass their turn, the game ends, and the winner is the player that colored more nodes.</p>

<p>You are the second player. If it is possible to choose such a <code>y</code> to ensure you win the game, return <code>true</code>. If it is not possible, return <code>false</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/08/01/1480-binary-tree-coloring-game.png" style="width: 500px; height: 310px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,4,5,6,7,8,9,10,11], n = 11, x = 3
<strong>Output:</strong> true
<strong>Explanation: </strong>The second player can choose the node with value 2.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = [1,2,3], n = 3, x = 1
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is <code>n</code>.</li>
	<li><code>1 &lt;= x &lt;= n &lt;= 100</code></li>
	<li><code>n</code> is odd.</li>
	<li>1 &lt;= Node.val &lt;= n</li>
	<li>All the values of the tree are <strong>unique</strong>.</li>
</ul>

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def btreeGameWinningMove(self, root: Optional[TreeNode], n: int, x: int) -> bool:
        node = self.findNode(root, x)
        left = self.count(node.left)
        right = self.count(node.right)
        choseParent = n - 1 - left - right
        print(choseParent, left, right)
        if choseParent > 1 + left + right:
            return True
        elif left > choseParent + 1 + right or right > choseParent + 1 + left:
            return True
        else:
            return False
    def findNode(self, root, x):
        if not root:
            return None
        elif root.val == x:
            return root
        left = self.findNode(root.left, x)
        right = self.findNode(root.right, x)
        if left:
            return left
        if right:
            return right

    def count(self, node):
        if node:
            return 1 + self.count(node.left) + self.count(node.right)
        else:
            return 0
        
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