# 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 :

                        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
int tmp = target;
while (tmp != 1) {
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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