**Binary Tree Width - Amazon Top Interview Questions**

### Problem Statement :

Given a binary tree root, return the maximum width of any level in the tree. The width of a level is the number of nodes that can fit between the leftmost node and the rightmost node. Constraints 1 ≤ n ≤ 100,000 where n is the number of nodes in root Example 1 Input root = [0, [1, [3, null, null], null], [2, [4, [5, null, null], [6, null, null]], null]] Output 3 Explanation The maximum width is 3 since between nodes 3 and 4, we can fit total of 3 nodes: [3, null, 4] Example 2 Input root = [0, [1, [3, null, null], null], [2, null, [4, null, null]]] Output 4 Explanation We can fit 4 nodes between 3 and 4. Example 3 Input root = [0, null, null] Output 1

### Solution :

` ````
Solution in C++ :
int solve(Tree* root) {
queue<pair<Tree*, int>> q{{{root, 1}}};
int ans = 1;
while (!q.empty()) {
int sz = q.size(), l = INT_MAX, r = INT_MIN;
while (sz-- > 0) {
auto& p = q.front();
l = min(l, p.second), r = max(r, p.second);
if (p.first->left) q.push({p.first->left, 2 * p.second - 1});
if (p.first->right) q.push({p.first->right, 2 * p.second});
q.pop();
}
ans = max(ans, r - l + 1);
}
return ans;
}
```

` ````
Solution in Java :
import java.util.*;
/**
* public class Tree {
* int val;
* Tree left;
* Tree right;
* }
*/
class Solution {
public int solve(Tree root) {
LinkedList<Pair<Tree, Integer>> q = new LinkedList();
q.addLast(new Pair(root, 0));
int res = 0;
while (q.size() > 0) {
int size = q.size();
// compare leftmost and rightmost
Pair<Tree, Integer> left = q.getFirst();
Pair<Tree, Integer> right = q.getLast();
int left_column = left.getValue();
int right_column = right.getValue();
res = Math.max(res, right_column - left_column + 1);
for (int i = 0; i < size; i++) { // standard BFS
Pair<Tree, Integer> cur = q.removeFirst();
Tree cur_node = cur.getKey();
int cur_column = cur.getValue();
if (cur_node.left != null) {
q.addLast(new Pair(cur_node.left, 2 * cur_column));
}
if (cur_node.right != null) {
q.addLast(new Pair(cur_node.right, 2 * cur_column + 1));
}
}
}
return res;
}
}
```

` ````
Solution in Python :
from typing import Deque
class Solution:
def solve(self, root):
max_width = 0
queue: Deque[list[tuple[Tree, int]]] = deque([(root, 1)]) if root else deque()
while queue:
max_width = max(max_width, queue[-1][1] - queue[0][1] + 1)
length = len(queue)
for _ in range(length):
node, index = queue.popleft()
for child, new_index in ((node.left, 2 * index), (node.right, 2 * index + 1)):
if child:
queue.append((child, new_index))
return max_width
```

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