human
Solve in Python:
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

 
Constraints:

TreeNode.val == -1
The height of the binary tree is less than or equal to 20
The total number of nodes is between [1, 10^4]
Total calls of find() is between [1, 10^4]
0 <= target <= 10^6
assistant
#Here is the Python code you requested.

# 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: TreeNode):
        
        self.root = root
        
        def helper(root, val):

            if not root:
                return
            
            root.val = val
            helper(root.left, root.val*2+1)
            helper(root.right, root.val*2+2)
                
        
        helper(self.root, 0)
        #print(self.root.val)
        return
            
    def find(self, target: int) -> bool:
        #遍历, 太慢
        #def bianli(root, target):
        #    if not root:
        #        return False
        #    
        #    if root.val == target:
        #        return True
        #    
        #    res = bianli(root.left,target) or bianli(root.right,target)
        #    return res
        #
        #return bianli(self.root, target)
        
        #因为这个tree的点不重复，所以可以找出唯一路径到target
        #那么将这个路径走一遍，如果不通就说明找不到
        
        def findpath(target):
            path = [target]
            while target:
                diff = 1 if target%2 else 2
                target = int((target - diff)/2)
                path.append(target)
                
            return path[::-1]
                
            
        def trav(node, depth, path):
            if depth> len(path):
                return False
            else:
                if not node:
                    return False
                if node.val == path[-1]:
                    return True
                #print(node.val)
                if node.val == path[depth]:
                    res = trav(node.left, depth+1, path) or trav(node.right, depth+1,path)
                    return res
                    
        path = findpath(target)
        #print(path)
        return trav(self.root, 0, path)
            
                
            
            
            
            
        
 ...