Search in a Virtually Complete Binary Tree- Google Top Interview Questions


Problem Statement :


Consider a complete binary tree of n nodes whose values are 1 to n. The root has value of 1, its left child is 2 and its right child is 3. In general, nodes' values are labelled 1 to n in level order traversal.

You are given a binary tree root and an integer target. Given that the root was originally a complete binary tree whose values were labelled as described above, and that some of the subtrees were deleted, return whether target exists in root.

Bonus: solve in \mathcal{O}(h)O(h) time where h is the height of the tree.

Constraints

1 ≤ n ≤ 100,000 where n is the number of nodes in root

Example 1

Input

root = [1, [2, [4, null, null], null], [3, [6, null, null], [7, null, null]]]
target = 6

Output

True

Explanation

6 exists in this tree.

Example 2

Input

root = [1, [2, [4, null, null], null], [3, [6, null, null], null]]

target = 7

Output

False

Explanation

7 does not exist in this tree.


Solution :



title-img 




                        Solution in C++ :

*/
bool check(Tree* root, vector<int>& path) {
    while (root) {
        if (root->val != path.back()) return false;
        path.pop_back();
        if (path.empty()) return true;
        if (path.back() % 2)
            root = root->right;
        else
            root = root->left;
    }
    return false;
}

bool solve(Tree* root, int target) {
    if (target < 1) return false;
    vector<int> path{target};
    for (int i = target; i != 1;) {
        i /= 2;
        path.push_back(i);
    }
    return check(root, path);
}
                    


                        Solution in Java :

import java.util.*;

/**
 * public class Tree {
 *   int val;
 *   Tree left;
 *   Tree right;
 * }
 */
class Solution {
    public boolean solve(Tree root, int target) {
        if (target <= 0)
            return false;

        // we can construct the path to target
        LinkedList<Integer> path = new LinkedList();
        int tmp = target;
        while (tmp != 1) {
            path.addFirst(tmp);
            tmp = tmp / 2;
        }
        Tree trav = root;
        while (path.size() > 0) {
            int num = path.removeFirst();
            if (num % 2 == 0) {
                if (trav.left == null)
                    return false;
                trav = trav.left;
            } else {
                if (trav.right == null)
                    return false;
                trav = trav.right;
            }
        }
        return true;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, root, target):
        if target == 0:
            return False
        path = []
        while target > 1:
            if target % 2:
                path.append(1)
            else:
                path.append(-1)
            target //= 2
        for x in path[::-1]:
            if x == -1:
                if root.left is None:
                    return False
                root = root.left
            else:
                if root.right is None:
                    return False
                root = root.right
        return True


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