1387. Find Elements In A Contaminated Binary Tree¶

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1387. Find Elements in a Contaminated Binary Tree

Medium

Given a binary tree with the following rules:

1. root.val == 0
2. For any treeNode:
   1. If treeNode.val has a value x and treeNode.left != null, then treeNode.left.val == 2 * x + 1
   2. If treeNode.val has a value 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, 104]
• Total calls of find() is between [1, 104]
• 0 <= target <= 106

Solution¶

/**
 * 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 TreeNode new_root;
    private HashSet<Integer> set;
    public FindElements(TreeNode root) {
        Queue<TreeNode> q = new LinkedList<>();
        Queue<TreeNode> q1 = new LinkedList<>();
        new_root = new TreeNode(0);
        set = new HashSet<>();
        q.offer(new_root);
        q1.offer(root);
        while (q.size() > 0) {
            int len = q.size();
            for (int i = 0; i < len; i++) {
                TreeNode current = q.peek();
                int val = q.peek().val;
                set.add(val);
                if (q1.peek().left != null) {
                    current.left = new TreeNode(val * 2 + 1);
                    q.offer(current.left);
                    q1.offer(q1.peek().left);
                    set.add(val * 2 + 1);
                }
                if (q1.peek().right != null) {
                    current.right = new TreeNode(val * 2 + 2);
                    q.offer(current.right);
                    q1.offer(q1.peek().right);
                    set.add(val * 2 + 2);
                }
                q.poll();
                q1.poll();
            }
        }
    }
    public boolean find(int target) {
        return set.contains(target);
    }
}

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

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here