# 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) {
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) {
}
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```
```

## Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio

You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis

## Merge two sorted linked lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C

## Get Node Value

This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t

## Delete duplicate-value nodes from a sorted linked list

This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -