Formatted question description: https://leetcode.ca/all/1261.html

1261. Find Elements in a Contaminated Binary Tree

Level

Medium

Description

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 true if the target value exists in the recovered binary tree.

Example 1:

Image text

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:

Image text

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:

Image text

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

In the constructor, recover the tree using breadth first search. The root’s value is 0. For each node that is not a leaf, obtain its two children, and if a child is not null, recover the child’s value.

In the method boolean find(int target), first obtain the path from the root to the target, and then check whether there exists a node with value target. The path can be obtained using the relations between the parent’s value and its children’s values. To find the node, also use the relations.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class FindElements {
    TreeNode root;

    public FindElements(TreeNode root) {
        if (root != null) {
            breadthFirstSearch(root);
            this.root = root;
        }
    }

    public boolean find(int target) {
        if (root == null)
            return false;
        Stack<Integer> stack = new Stack<Integer>();
        int value = target;
        while (value > 0) {
            stack.push(value);
            value = (value - 1) / 2;
        }
        TreeNode node = root;
        while (node != null && !stack.isEmpty()) {
            int nextValue = stack.pop();
            if (nextValue == 2 * value + 1)
                node = node.left;
            else if (nextValue == 2 * value + 2)
                node = node.right;
            if (node != null && nextValue == target)
                return true;
            else
                value = nextValue;
        }
        return false;
    }

    private void breadthFirstSearch(TreeNode root) {
        root.val = 0;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            int val = node.val;
            TreeNode left = node.left, right = node.right;
            if (left != null) {
                left.val = 2 * val + 1;
                queue.offer(left);
            }
            if (right != null) {
                right.val = 2 * val + 2;
                queue.offer(right);
            }
        }
    }
}

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

// OJ: https://leetcode.com/problems/find-elements-in-a-contaminated-binary-tree/
// Time:
//      FindElements: O(N)
//      find: O(1)
// Space: O(N)
class FindElements {
    unordered_set<int> s;
    void recover(TreeNode *root, int val) {
        if (!root) return;
        root->val = val;
        s.insert(val);
        recover(root->left, 2 * val + 1);
        recover(root->right, 2 * val + 2);
    }
public:
    FindElements(TreeNode* root) {
        recover(root, 0);
    }
    bool find(int target) {
        return s.count(target);
    }
};

# 1261. Find Elements in a Contaminated Binary Tree
# https://leetcode.com/problems/find-elements-in-a-contaminated-binary-tree/

# 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):
        root.val = 0
        self.s = set()
        self.build(root)
        
    def find(self, target: int) -> bool:
        return target in self.s
    
    def build(self, root):
        if not root: return
        
        if root.left:
            root.left.val = root.val*2 + 1
            self.s.add(root.left.val)
            self.build(root.left)
        
        if root.right:
            root.right.val = root.val*2 + 2
            self.s.add(root.right.val)
            self.build(root.right)
        


# Your FindElements object will be instantiated and called as such:
# obj = FindElements(root)
# param_1 = obj.find(target)

1261 - Find Elements in a Contaminated Binary Tree

1261. Find Elements in a Contaminated Binary Tree

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