Level Order Traversal - Amazon Top Interview Questions

Problem Statement :

Given a binary tree root return a level order traversal of the node values.


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

Example 1


root = [0, [5, null, null], [9, [1, [4, null, null], [2, null, null]], [3, null, null]]]


[0, 5, 9, 1, 3, 4, 2]

Example 2


root = [0, [1, [2, [3, null, null], null], null], null]


[0, 1, 2, 3]

Example 3


root = [0, null, [1, null, [2, null, [3, null, null]]]]


[0, 1, 2, 3]

Solution :


                        Solution in C++ :

vector<int> solve(Tree* root) {
    vector<int> res;
    if (root == nullptr) return {};
    queue<Tree*> q;
    while (!q.empty()) {
        Tree* n = q.front();
        if (n->left) q.push(n->left);
        if (n->right) q.push(n->right);
    return res;

                        Solution in Java :

import java.util.*;

 * public class Tree {
 *   int val;
 *   Tree left;
 *   Tree right;
 * }
class Solution {
    public int[] solve(Tree root) {
        if (root == null)
            return new int[0];
        ArrayList<Integer> list = new ArrayList<>();
        Queue<Tree> queue = new LinkedList<>();
        while (!queue.isEmpty()) {
            Tree pop = queue.poll();
            if (pop.left != null) {
            if (pop.right != null) {

        return convertIntegers(list);

    public int[] convertIntegers(List<Integer> integers) {
        int[] ret = new int[integers.size()];
        Iterator<Integer> iterator = integers.iterator();
        for (int i = 0; i < ret.length; i++) {
            ret[i] = iterator.next().intValue();
        return ret;

                        Solution in Python : 
class Solution:
    def solve(self, root):
        q = [root]
        ans = []
        for node in q:
            if node.left:
            if node.right:
        return ans

        # ans = []
        # if root:
        #     if a==True:
        #         ans.append(root.val)
        #     if root.left:
        #         ans.append(root.left.val)
        #     if root.right:
        #         ans.append(root.right.val)
        #     ans += self.solve(root.left,False)
        #     ans += self.solve(root.right,False)

        # return ans

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