<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.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool btreeGameWinningMove(TreeNode* root, int n, int x) {
        stack<TreeNode*> stk;
        map<TreeNode*, int> visit;
        stk.push(root);
        int p1left, p1right, player1;
        while(!stk.empty()){
            TreeNode* cur = stk.top();
            if((!cur->left || visit[cur->left]) && (!cur->right || visit[cur->right])){
                int left = (cur->left == NULL) ? 0 : cur->left->val;
                int right = (cur->right == NULL) ? 0 : cur->right->val;
                if(cur->val == x){
                    player1 = left + right + 1;
                    p1left = left;
                    p1right = right;
                }
                cur->val = left + right + 1;
                visit[cur] = 1;
                stk.pop();
            }
            else if(cur->left && !visit[cur->left]){
                stk.push(cur->left);
            }
            else if(cur->right && !visit[cur->right]){
                stk.push(cur->right);
            }
        }
        return max(max(p1left, p1right), n - p1left - p1right - 1) > n / 2;
    }
};
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