Left Side View of a Tree - Amazon Top Interview Questions
Problem Statement :
Given a binary tree root, return the leftmost node's value on each level of the tree. Constraints n ≤ 100,000 where n is the number of nodes in root Example 1 Input root = [0, [5, null, null], [2, null, [1, null, null]]] Output [0, 5, 1]
Solution :
Solution in C++ :
vector<int> solve(Tree* root) {
vector<int> res;
queue<Tree*> q;
q.push(root);
while (q.size() != 0) {
int size = q.size();
for (int i = 0; i < size; i++) {
Tree* top = q.front();
q.pop();
if (i == 0) res.push_back(top->val);
if (top->left != NULL) q.push(top->left);
if (top->right != NULL) q.push(top->right);
}
}
return res;
}
Solution in Java :
import java.util.*;
/**
* public class Tree {
* int val;
* Tree left;
* Tree right;
* }
*/
class Solution {
List<Integer> list = new ArrayList<>();
public int[] solve(Tree root) {
helper(root, 0);
// converts List<Integer> to int[] array
return list.stream().mapToInt(Integer::intValue).toArray();
}
private void helper(Tree root, int level) {
if (root != null) {
if (list.size() == level) {
list.add(root.val);
}
helper(root.left, level + 1);
helper(root.right, level + 1);
}
}
}
Solution in Python :
class Solution:
def solve(self, root):
self.hash_map = set()
self.output = []
self.traverse(root, 0)
return self.output
def traverse(self, node, level):
if not node:
return
if level not in self.hash_map:
self.hash_map.add(level)
self.output.append(node.val)
self.traverse(node.left, level + 1)
self.traverse(node.right, level + 1)
View More Similar Problems
Array-DS
An array is a type of data structure that stores elements of the same type in a contiguous block of memory. In an array, A, of size N, each memory location has some unique index, i (where 0<=i<N), that can be referenced as A[i] or Ai. Reverse an array of integers. Note: If you've already solved our C++ domain's Arrays Introduction challenge, you may want to skip this. Example: A=[1,2,3
View Solution →2D Array-DS
Given a 6*6 2D Array, arr: 1 1 1 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 An hourglass in A is a subset of values with indices falling in this pattern in arr's graphical representation: a b c d e f g There are 16 hourglasses in arr. An hourglass sum is the sum of an hourglass' values. Calculate the hourglass sum for every hourglass in arr, then print t
View Solution →Dynamic Array
Create a list, seqList, of n empty sequences, where each sequence is indexed from 0 to n-1. The elements within each of the n sequences also use 0-indexing. Create an integer, lastAnswer, and initialize it to 0. There are 2 types of queries that can be performed on the list of sequences: 1. Query: 1 x y a. Find the sequence, seq, at index ((x xor lastAnswer)%n) in seqList.
View Solution →Left Rotation
A left rotation operation on an array of size n shifts each of the array's elements 1 unit to the left. Given an integer, d, rotate the array that many steps left and return the result. Example: d=2 arr=[1,2,3,4,5] After 2 rotations, arr'=[3,4,5,1,2]. Function Description: Complete the rotateLeft function in the editor below. rotateLeft has the following parameters: 1. int d
View Solution →Sparse Arrays
There is a collection of input strings and a collection of query strings. For each query string, determine how many times it occurs in the list of input strings. Return an array of the results. Example: strings=['ab', 'ab', 'abc'] queries=['ab', 'abc', 'bc'] There are instances of 'ab', 1 of 'abc' and 0 of 'bc'. For each query, add an element to the return array, results=[2,1,0]. Fun
View Solution →Array Manipulation
Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array. Example: n=10 queries=[[1,5,3], [4,8,7], [6,9,1]] Queries are interpreted as follows: a b k 1 5 3 4 8 7 6 9 1 Add the valu
View Solution →