1261. Find Elements in a Contaminated Binary Tree

Problem

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.

Implement the FindElements class:

• FindElements(TreeNode* root) Initializes the object with a contaminated binary tree and recovers it.

• bool find(int target) Returns true 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

Solution (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 {
    private final HashMap<Integer, Integer> map = new HashMap<>();

    public FindElements(TreeNode root) {
        helper(root, 0);
    }

    private void helper(TreeNode root, int x) {
        if (root == null) {
            return;
        }
        root.val = x;
        map.put(x, 0);
        helper(root.left, 2 * x + 1);
        helper(root.right, 2 * x + 2);
    }

    public boolean find(int target) {
        return map.containsKey(target);
    }
}

/**
 * Your FindElements object will be instantiated and called as such:
 * FindElements obj = new FindElements(root);
 * boolean param_1 = obj.find(target);
 */

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).