Level Order Traversal - Amazon Top Interview Questions


Problem Statement :


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

Constraints

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

Example 1

Input

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

Output

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

Example 2

Input

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

Output

[0, 1, 2, 3]

Example 3

Input

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

Output

[0, 1, 2, 3]


Solution :



title-img 




                        Solution in C++ :

vector<int> solve(Tree* root) {
    vector<int> res;
    if (root == nullptr) return {};
    queue<Tree*> q;
    q.push(root);
    while (!q.empty()) {
        Tree* n = q.front();
        q.pop();
        res.push_back(n->val);
        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<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            Tree pop = queue.poll();
            list.add(pop.val);
            if (pop.left != null) {
                queue.offer(pop.left);
            }
            if (pop.right != null) {
                queue.offer(pop.right);
            }
        }

        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:
                q.append(node.left)
            if node.right:
                q.append(node.right)
            ans.append(node.val)
        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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