# 1261. 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

# Solution

Approach 1: use a set to hold all values.

# Code (Python)

Approach 1:

class FindElements:

    def __init__(self, root: TreeNode):
        self.vals = set()
        root.val = 0
        stack = [root]
        while stack:
            node = stack.pop()
            self.vals.add(node.val)
            if node.left:
                node.left.val = 2 * node.val + 1
                stack.append(node.left)
            if node.right:
                node.right.val = 2 * node.val + 2
                stack.append(node.right)

    def find(self, target: int) -> bool:
        return target in self.vals

# Code (C++)

Approach 1:

Approach 2: