Python DFS Solution
If we observe that there are three "optimal" choices for \'y\'. \'y\' could either choose \'x\'\'s parent, x\'s left child or x\'s right child. Then the final o
If we observe that there are three "optimal" choices for \'y\'. \'y\' could either choose \'x\'\'s parent, x\'s left child or x\'s right child. Then the final optimal choice for \'y\' is the optimal among those three. If any choice result wining of \'y\', then \'y\' will win. Otherwise, \'y\' will loose.\n\n\nBelow is my python implementation\n\n```python\n\n# Definition for a binary tree node.\n# class TreeNode:\n#     def __init__(self, x):\n#         self.val = x\n#         self.left = None\n#         self.right = None\n\nclass Solution:\n    def btreeGameWinningMove(self, root: TreeNode, n: int, x: int) -> bool:\n        \n        \'\'\'\n        if choose x\'s subtree, better choose the root of its larger subtree\n        if choose x\'s ancestors, better choose x\'s parent.\n        \'\'\'\n        childcount = collections.defaultdict(int)\n        leftchildren = {}\n        rightchildren = {}\n        parents = {}\n        \n        def dfs(root:TreeNode, parent:TreeNode)->int: #return a node\'s child node count including itself\n            if root == None:\n                return 0\n            left = dfs(root.left, root)\n            right = dfs(root.right, root)\n            childcount[root.val] = left + right + 1\n            parents[root.val] = parent\n            leftchildren[root.val] = root.left\n            rightchildren[root.val] = root.right\n            return left + right + 1\n        \n        dfs(root, None)\n        #case1\n        if parents[x] != None:\n            m = n - childcount[x]\n            if m > n // 2:\n                return True\n        #case2\n        if leftchildren[x] != None:\n            m = childcount[leftchildren[x].val]\n            if m > n // 2:\n                return True\n        #case3\n        if rightchildren[x] != None:\n            m = childcount[rightchildren[x].val]\n            if m > n // 2:\n                return True\n        return False\n        \n```
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