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Level Order Traversal

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Problem statement

You are given a ‘Binary Tree’.


Return the level-order traversal of the Binary Tree.


Example:
Input: Consider the following Binary Tree:

Example

Output: 
Following is the level-order traversal of the given Binary Tree: [1, 2, 3, 5, 6, 4]


Detailed explanation ( Input/output format, Notes, Images )
Input Format:
The only line contains elements in the level order form. The line consists of values of nodes separated by a single space. In case a node is null, we take -1 in its place.

For example, the input for the tree depicted in the below image will be:

alt text

1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1

Explanation :
Level 1 :
The root node of the tree is 1

Level 2 :
Left child of 1 = 2
Right child of 1 = 3

Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6

Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)

Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)

The first not-null node(of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.

The input ends when all nodes at the last level are null(-1).

The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:

1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1


Output Format:
Return an array representing the level-order traversal of the given binary tree.


Note :
You do not need to print anything; it has already been taken care of. Just implement the given function.
Sample Input 1:
1 2 3 5 4 6 7 -1 -1 -1 -1 -1 -1 -1 -1


Sample Output 1:
1 2 3 5 4 6 7


Explanation of Sample Input 1:
The Binary Tree given in the input is as follows:

Sample1

Sample Input 2:
5 6 10 2 3 -1 -1 -1 -1 -1 9 -1 -1


Sample Output 2:
5 6 10 2 3 9


Explanation of Sample Input 2:
The Binary Tree given in the input is as follows:

Sample2

Expected Time Complexity:
Try to do this in O(n).


Constraints:
1 <= n <= 100000
where 'n' is the number of nodes in the binary tree.

Time Limit: 1 sec
Hint

Use queue.

Approaches (1)
Breadth-First Search

Approach:

While traversing the binary tree in a level order way, we traverse the tree level by level, from left to right. This is essentially a breadth-first traversal of the binary tree. We will use a queue to maintain the order of the nodes that we need to traverse. Initially the queue will contain only the root of the tree. Then for each node, we will do the following:

  • Pop the node in the front of the queue.
  • Add its left child to the end of the queue.
  • Add its right child to the end of the queue.
  • Append the value of the current node to the path.


 

The steps are as follows:

function levelOrder(TreeNode<int>* root)

  1. Initialise empty array path
  2. Initialise empty queue ‘q’
  3. q.push(root)
  4. while(!q.empty())
    1. cur = q.pop()
    2. Push value of current node to the path
    3. if(cur->left)
      1. q.push(cur->left)
    4. if(cur->right)
      1. q.push(cur->right)
  5. return path

   

Time Complexity

Time Complexity:

O(N), where ‘N’ is the number of nodes in the tree.


 

We visit all the nodes present in the binary tree.


 

Hence, the time complexity is O(N).

Space Complexity

O(W), where ‘W’ is the maximum width of the binary tree.


 

At most ‘W’ nodes will be stored in the queue while traversing the tree.


Hence, the space complexity is O(W).
 

Code Solution
(100% EXP penalty)
Level Order Traversal
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Interview problems

C++||Easy Solution||Level Order Traversal||

vector<int> levelOrder(TreeNode<int> * root){

   

    queue<TreeNode<int>*>q;

    q.push(root);

    vector<int>ans;

    while(!q.empty())

    {

        TreeNode<int>*  temp=q.front();

        q.pop();

        ans.push_back(temp->data);

        

        if(temp->left){

            q.push(temp->left);

        }

        if(temp->right)

        {

            q.push(temp->right);

        }

    }

    return  ans;

}

 

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Interview problems

python easy solution

from collections import deque
from typing import List

def levelOrder(root) -> List[int]:
    q=deque()
    v=[]
    q.append(root)

    while q:
        temp=q.popleft()
        v.append(temp.data)
        
        if temp.left:
            q.append(temp.left)

        if temp.right:
            q.append(temp.right)
    
    return v
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Interview problems

Level Order Traversal C++ Easy solution

https://github.com/yashswag22/Striver-A-to-Z-DSA-sheet  < = Coding ninjas Striver sheet solution available.

vector<int> levelOrder(TreeNode<int> * root){
    vector<int>ans;
    queue<TreeNode<int>*>q;
    q.push(root);
    while(!q.empty()){
        
        TreeNode<int>* temp = q.front();
        q.pop();
        ans.push_back(temp->data);

        if(temp->left != nullptr)
        q.push(temp->left);
        if(temp->right != nullptr)
        q.push(temp->right);

    }

    return ans;
}

Reverse Level Order Traversal

Zigzag Binary Tree Traversal

Level Order Traversal

Easy

+1 more
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Java solution

import java.util.*;

public class Solution {

    public static List<Integer> levelOrder(TreeNode<Integer> root){

        // Write your code here.

        Queue<TreeNode<Integer>> queue=new LinkedList<TreeNode<Integer>>();

        List<Integer> l=new ArrayList<>();

        queue.add(root);

        if(root==null){

            return l;

        }

        while(!queue.isEmpty()){

            TreeNode<Integer> node = queue.poll();

            if (node != null) {

                l.add(node.data);

                queue.add(node.left);

                queue.add(node.right);

 

            }

        }

        return l;

    }

}

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Interview problems

C++ || Easy Solution || All tcs passed 🔥

Approach:

  1. Initialize an empty vector level to store the nodes in level-order and a queue q to perform the traversal.
  2. Enqueue the root node into the queue.
  3. While the queue is not empty, repeat the following steps:
    • Get the current size of the queue (represents the number of nodes at the current level).
    • Iterate over each node at the current level:
      • Dequeue the front node of the queue.
      • Enqueue its left and right children (if they exist) into the queue.
      • Add the data of the dequeued node to the level vector.
  4. Once the traversal is complete, return the level vector containing the nodes in level-order

 

Time Complexity:  O(n), Since we are visiting every node. Space Complexity: O(n) but we should not consider this because function expects us to return an array.

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Java Easy Solution

public class Solution {

    public static List<Integer> levelOrder(TreeNode<Integer> root){

        // Write your code here.

 

        Queue<TreeNode<Integer>> q = new LinkedList<>();

        List<Integer> list = new ArrayList<>(); 

        q.add(root);

 

        while (!q.isEmpty()) {

            TreeNode<Integer> node = q.poll();

 

            if (node != null) {

                list.add(node.data);

                q.add(node.left);

                q.add(node.right);

            }

        }

 

        return list;

    }

}

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Interview problems

easy cpp

/************************************************************

 

    Following is the TreeNode class structure

 

    template <typename T>

    class TreeNode {

       public:

        T data;

        TreeNode<T> *left;

        TreeNode<T> *right;

 

        TreeNode(T data) {

            this->data = data;

            left = NULL;

            right = NULL;

        }

    };

 

************************************************************/

 

vector<int> levelOrder(TreeNode<int> * root){

    // Write your code here.

    vector<int> ans;

    if(root==NULL){

        return ans;

    }

    queue<TreeNode<int>*> q1;

    q1.push(root);

    while(!q1.empty()){

        int size=q1.size();

        for(int i=1;i<=size;i++){

            TreeNode<int> * Node=q1.front();

            q1.pop();

            if(Node->left!=NULL) q1.push(Node->left);

            if(Node->right!=NULL) q1.push(Node->right);

            ans.push_back(Node->data);

        } 

    }

    return ans;

 

}

 

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Interview problems

+ Strivers CPP Solution +

#include<queue>
vector<int> levelOrder(TreeNode<int> * root){
    queue<TreeNode<int>*>q;
    vector<int>ans;
    q.push(root);
    while(!q.empty()){
        root=q.front();
        if(root->left)q.push(root->left);
        if(root->right)q.push(root->right);
        q.pop();
        ans.push_back(root->data);
    }
    return ans;
}
# Please upvote if it helped # It would help me
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Interview problems

C++ || Easy to understand

void levelBT(TreeNode<int> * root, vector<int> &ans){

 

    if(root == NULL) return;

 

    queue<TreeNode<int>*> q;

    q.push(root);

 

    while(!q.empty()){

 

        TreeNode<int>* temp = q.front();

        q.pop();

 

        ans.push_back(temp->data);

 

        if(temp->left) q.push(temp->left);

 

        if(temp->right) q.push(temp->right);

 

    }

 

}

 

vector<int> levelOrder(TreeNode<int> * root){

    

    vector<int> ans;

    levelBT(root, ans);

    return ans;

 

}

 

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Easy Java Solution

public class Solution {
    public static List<Integer> levelOrder(TreeNode<Integer> root){
        List<Integer> ans = new ArrayList<>();
        Queue<TreeNode<Integer>> q = new LinkedList<>();


        q.add(root);


        while(!q.isEmpty()){
          TreeNode<Integer> node = q.poll();
            ans.add(node.data);


           if(node.left != null){
            q.add(node.left);
            } 
           if(node.right != null){
            q.add(node.right);
        }
    }
        return ans;
    }
}
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