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 :



title-img




                        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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