Description
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
分析
题目的意思是：给出一颗二叉树，和两个player，其中player的位置是x，问y放在哪个位置能够使y赢，两个player可以在未填充的区域填上自己的颜色，然后在自己的颜色区域范围内移动，当两个player都没有再填充新的区域了之后，游戏结束，谁填充的颜色多就算谁赢。
- y的位置有三种选择，在x的左子树，右子树和y的父结点。这里有个规律，y必须放在未涂色个数多的一侧，并且结点的个数要大于n/2，并且放在与他相邻的位置就行了。这样能够锁定下面的子树。这道题需要找规律，感觉还是挺麻烦的，如果能够知道要针对这三种情况，并且y所在的其中一种情况必须大于n/2就能做出来了。
代码
# 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 solve(self,root,x):
if(root is None):
return 0
l=self.solve(root.left,x)
r=self.solve(root.right,x)
if(root.val==x):
self.res=[l,r]
return l+r+1
def btreeGameWinningMove(self, root: TreeNode, n: int, x: int) -> bool:
self.res=[0,0]
self.solve(root,x)
t=max(max(self.res),n-sum(self.res)-1)
if(n/2<t):
return True
return False