leetcode1261 Find Elements in a Contaminated Binary Tree

 """
 Given a binary tree with the following rules:
     root.val == 0
     If treeNode.val == x and treeNode.left != null, then treeNode.left.val == 2 * x + 1
     If treeNode.val == x and treeNode.right != null, then treeNode.right.val == 2 * x + 2
 Now the binary tree is contaminated, which means all treeNode.val have been changed to -1.
 You need to first recover the binary tree and then implement the FindElements class:
     FindElements(TreeNode* root) Initializes the object with a contamined binary tree, you need to recover it first.
     bool find(int target) Return if the target value exists in the recovered binary tree.
 Example 1:
 Input
 ["FindElements","find","find"]
 [[[-1,null,-1]],[1],[2]]
 Output
 [null,false,true]
 Explanation
 FindElements findElements = new FindElements([-1,null,-1]);
 findElements.find(1); // return False
 findElements.find(2); // return True
 Example 2:
 Input
 ["FindElements","find","find","find"]
 [[[-1,-1,-1,-1,-1]],[1],[3],[5]]
 Output
 [null,true,true,false]
 Explanation
 FindElements findElements = new FindElements([-1,-1,-1,-1,-1]);
 findElements.find(1); // return True
 findElements.find(3); // return True
 findElements.find(5); // return False
 Example 3:
 Input
 ["FindElements","find","find","find","find"]
 [[[-1,null,-1,-1,null,-1]],[2],[3],[4],[5]]
 Output
 [null,true,false,false,true]
 Explanation
 FindElements findElements = new FindElements([-1,null,-1,-1,null,-1]);
 findElements.find(2); // return True
 findElements.find(3); // return False
 findElements.find(4); // return False
 findElements.find(5); // return True
 """
 class TreeNode:
     def __init__(self, x):
         self.val = x
         self.left = None
         self.right = None

 class FindElements:
     def __init__(self, root):
         self.array = []  # !!!保存结点出现的值,注意self. 私有成员
         if root != None:  # 调用递归函数
             self.recover_tree(root, 0)
     def recover_tree(self, root, v):  # 递归调用
         root.val = v
         self.array.append(v)
         if root.left != None:
             self.recover_tree(root.left, 2 * v + 1)
         if root.right != None:
             self.recover_tree(root.right, 2 * v + 2)
     def find(self, target: int) -> bool:
         return target in self.array  # 再结果数组里找