# Complete Binary Tree - Amazon Top Interview Questions

### Problem Statement :

```Given a binary tree root, return whether it's a complete binary tree.

Constraints

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

Example 1

Input

root = [0, [1, [1, null, null], [0, null, null]], [0, [1, null, null], [0, null, null]]]

Output

True

Example 2

Input

root = [0, [1, [1, null, null], [0, null, null]], [0, null, null]]

Output

True

Example 3

Input

root = [0, [1, null, [0, null, null]], [0, null, null]]

Output

False```

### Solution :

```                        ```Solution in C++ :

bool solve(Tree *root) {
// okay I read levels and immediately thought of level order traversal ;)
// but as soon as i reach a point without a child
// nobody after that should have a child!!!!!!!!

bool no_kids = false;
queue<Tree *> q;
if (root) q.push(root);

while (not q.empty()) {
Tree *cur = q.front();
q.pop();

if (cur->left)
if (no_kids)
return false;
else
q.push(cur->left);
else
no_kids = true;

if (cur->right)
if (no_kids)
return false;
else
q.push(cur->right);
else
no_kids = true;
}

return true;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public boolean solve(Tree root) {
return dfs(root, 1) == count;
}
int count = 0;
private int dfs(Tree root, int index) {
if (root == null)
return 0;
count++;
return Math.max(index, Math.max(dfs(root.left, index * 2), dfs(root.right, index * 2 + 1)));
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, root):
if root:
queue = deque([root])
else:
return True
can_child = True

while queue:
node = queue.popleft()

if can_child:
if node.left:
queue.append(node.left)
else:
can_child = False
else:
if node.left:
return False

if can_child:
if node.right:
queue.append(node.right)
else:
can_child = False
else:
if node.right:
return False
return True```
```

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