给出一个满足下述规则的二叉树：

1. root.val == 0
2. 如果 treeNode.val == x 且 treeNode.left != null，那么 treeNode.left.val == 2 * x + 1
3. 如果 treeNode.val == x 且 treeNode.right != null，那么 treeNode.right.val == 2 * x + 2

现在这个二叉树受到「污染」，所有的 treeNode.val 都变成了 -1。

请你先还原二叉树，然后实现 FindElements 类：

• FindElements(TreeNode* root) 用受污染的二叉树初始化对象，你需要先把它还原。
• bool find(int target) 判断目标值 target 是否存在于还原后的二叉树中并返回结果。

示例 1：

输入：
["FindElements","find","find"]
[[[-1,null,-1]],[1],[2]]
输出：
[null,false,true]
解释：
FindElements findElements = new FindElements([-1,null,-1]); 
findElements.find(1); // return False 
findElements.find(2); // return True 

示例 2：

输入：
["FindElements","find","find","find"]
[[[-1,-1,-1,-1,-1]],[1],[3],[5]]
输出：
[null,true,true,false]
解释：
FindElements findElements = new FindElements([-1,-1,-1,-1,-1]);
findElements.find(1); // return True
findElements.find(3); // return True
findElements.find(5); // return False

示例 3：

输入：
["FindElements","find","find","find","find"]
[[[-1,null,-1,-1,null,-1]],[2],[3],[4],[5]]
输出：
[null,true,false,false,true]
解释：
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

提示：

• TreeNode.val == -1
• 二叉树的高度不超过 20
• 节点的总数在 [1, 10^4] 之间
• 调用 find() 的总次数在 [1, 10^4] 之间
• 0 <= target <= 10^6

问题简要描述：还原二叉树并实现 FindElements 类 

Java

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class FindElements {Set<Integer> set = new HashSet<>();public FindElements(TreeNode root) {root.val = 0;dfs(root);}public boolean find(int target) {return set.contains(target);}void dfs(TreeNode root) {set.add(root.val);if (root.left != null) {root.left.val = 2 * root.val + 1;dfs(root.left);}if (root.right != null) {root.right.val = 2 * root.val + 2;dfs(root.right);}}
}/*** Your FindElements object will be instantiated and called as such:* FindElements obj = new FindElements(root);* boolean param_1 = obj.find(target);*/

 Python3

# 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 FindElements:def __init__(self, root: Optional[TreeNode]):def dfs(root: Optional[TreeNode]):self.s.add(root.val)if root.left:root.left.val = root.val * 2 + 1dfs(root.left)if root.right:root.right.val = root.val * 2 + 2dfs(root.right)self.s = set()root.val = 0dfs(root)def find(self, target: int) -> bool:return target in self.s        # Your FindElements object will be instantiated and called as such:
# obj = FindElements(root)
# param_1 = obj.find(target)

TypeScript

/*** Definition for a binary tree node.* class TreeNode {*     val: number*     left: TreeNode | null*     right: TreeNode | null*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {*         this.val = (val===undefined ? 0 : val)*         this.left = (left===undefined ? null : left)*         this.right = (right===undefined ? null : right)*     }* }*/class FindElements {private s = new Set();constructor(root: TreeNode | null) {const dfs = (root: TreeNode | null) => {this.s.add(root.val)if (root.left != null) {root.left.val = root.val * 2 + 1;dfs(root.left);}if (root.right != null) {root.right.val = root.val * 2 + 2;dfs(root.right);}}root.val = 0;dfs(root);        }find(target: number): boolean {return this.s.has(target);    }
}/*** Your FindElements object will be instantiated and called as such:* var obj = new FindElements(root)* var param_1 = obj.find(target)*/