1145. Binary Tree Coloring Game 树上博弈

博弈论在二叉树游戏中的应用

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本文探讨了一个基于二叉树的回合制游戏，其中两名玩家轮流选择节点进行染色，目标是在游戏结束时拥有更多的节点。文章提供了一个算法，用于确定第二名玩家是否有可能通过明智的选择来赢得游戏，这涉及到对树结构的深入理解和策略制定。

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

Initially, the first player names a value x with 1 <= x <= n, and the second player names a value y with 1 <= y <= n and y != x. The first player colors the node with value x red, and the second player colors the node with value y blue.

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 uncolored neighbor of the chosen node (either the left child, right child, or parent of the chosen node.)

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.

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

Example 1:

Input: root = [1,2,3,4,5,6,7,8,9,10,11], n = 11, x = 3
Output: true
Explanation: The second player can choose the node with value 2.

Constraints:

root is the root of a binary tree with n nodes and distinct node values from 1 to n.
n is odd.
1 <= x <= n <= 100

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先手占据某个点后，可以把树变成一个三叉树，后手试图占据三叉树中的最大即可，典型的脑筋急转弯题目

# 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: TreeNode, n: int, x: int) -> bool:
        def count_node(rt):
            nonlocal lsave,rsave,x
            if (rt == None):
                return 0
            xl = count_node(rt.left)
            xr = count_node(rt.right)
            if (rt.val == x):
                lsave = xl
                rsave = xr
            return xl+xr+1
        if (root == None):
            return True
        lsave,rsave = 0,0
        total = count_node(root)
        return max(lsave,rsave,total-lsave-rsave-1)*2>n