Connect Nodes at same Level (Level Order Traversal)

Write a function to connect all the adjacent nodes at the same level in a binary tree.

Example:

Input Tree
       A
      / \
     B   C
    / \   \
   D   E   F


Output Tree
       A--->NULL
      / \
     B-->C-->NULL
    / \   \
   D-->E-->F-->NULL

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

We have already discussed O(n^2) time and O approach in Connect nodes at same level as morris traversal in worst case can be O(n) and calling it to set right pointer can result in O(n^2) time complexity.

In this post, We have discussed Level Order Traversal with NULL markers which are needed to mark levels in tree.

C++

// Connect nodes at same level using level order 
// traversal. 
#include <bits/stdc++.h> 
using namespace std; 
  
struct Node { 
    int data; 
    struct Node* left, *right, *nextRight; 
}; 
  
// Sets nextRight of all nodes of a tree 
void connect(struct Node* root) 
{ 
    queue<Node*> q; 
    q.push(root); 
  
    // null marker to represent end of current level 
    q.push(NULL);  
  
    // Do Level order of tree using NULL markers 
    while (!q.empty()) { 
        Node *p = q.front(); 
        q.pop(); 
        if (p != NULL) { 
  
            // next element in queue represents next  
            // node at current Level  
            p->nextRight = q.front(); 
  
            // push left and right children of current 
            // node 
            if (p->left) 
                q.push(p->left);  
            if (p->right) 
                q.push(p->right); 
        }  
         
        // if queue is not empty, push NULL to mark  
        // nodes at this level are visited 
        else if (!q.empty())  
            q.push(NULL);  
    } 
} 
  
/* UTILITY FUNCTIONS */
/* Helper function that allocates a new node with the 
   given data and NULL left and right pointers. */
struct Node* newnode(int data) 
{ 
    struct Node* node = (struct Node*) 
                         malloc(sizeof(struct Node)); 
    node->data = data; 
    node->left = node->right = node->nextRight = NULL; 
    return (node); 
} 
  
/* Driver program to test above functions*/
int main() 
{ 
  
    /* Constructed binary tree is 
              10 
            /   \ 
          8      2 
        /         \ 
      3            90 
    */
    struct Node* root = newnode(10); 
    root->left = newnode(8); 
    root->right = newnode(2); 
    root->left->left = newnode(3); 
    root->right->right = newnode(90); 
  
    // Populates nextRight pointer in all nodes 
    connect(root); 
  
    // Let us check the values of nextRight pointers 
    printf("Following are populated nextRight pointers in \n"
     "the tree (-1 is printed if there is no nextRight) \n"); 
    printf("nextRight of %d is %d \n", root->data, 
     root->nextRight ? root->nextRight->data : -1); 
    printf("nextRight of %d is %d \n", root->left->data, 
     root->left->nextRight ? root->left->nextRight->data : -1); 
    printf("nextRight of %d is %d \n", root->right->data, 
     root->right->nextRight ? root->right->nextRight->data : -1); 
    printf("nextRight of %d is %d \n", root->left->left->data, 
     root->left->left->nextRight ? root->left->left->nextRight->data : -1); 
    printf("nextRight of %d is %d \n", root->right->right->data, 
     root->right->right->nextRight ? root->right->right->nextRight->data : -1); 
    return 0; 
} 

Java

// Connect nodes at same level using level order 
// traversal. 
import java.util.LinkedList; 
import java.util.Queue; 
public class Connect_node_same_level { 
      
    // Node class 
    static class Node { 
        int data; 
        Node left, right, nextRight; 
        Node(int data){ 
            this.data = data; 
            left = null; 
            right = null; 
            nextRight = null; 
        } 
    }; 
       
    // Sets nextRight of all nodes of a tree 
    static void connect(Node root) 
    { 
        Queue<Node> q = new LinkedList<Node>();  
        q.add(root); 
       
        // null marker to represent end of current level 
        q.add(null);  
       
        // Do Level order of tree using NULL markers 
        while (!q.isEmpty()) { 
            Node p = q.peek(); 
            q.remove(); 
            if (p != null) { 
       
                // next element in queue represents next  
                // node at current Level  
                p.nextRight = q.peek(); 
       
                // push left and right children of current 
                // node 
                if (p.left != null) 
                    q.add(p.left);  
                if (p.right != null) 
                    q.add(p.right); 
            }  
              
            // if queue is not empty, push NULL to mark  
            // nodes at this level are visited 
            else if (!q.isEmpty())  
                q.add(null);  
        } 
    } 
       
    /* Driver program to test above functions*/
    public static void main(String args[]) 
    { 
       
        /* Constructed binary tree is 
                  10 
                /   \ 
              8      2 
            /         \ 
          3            90 
        */
        Node root = new Node(10); 
        root.left = new Node(8); 
        root.right = new Node(2); 
        root.left.left = new Node(3); 
        root.right.right = new Node(90); 
       
        // Populates nextRight pointer in all nodes 
        connect(root); 
       
        // Let us check the values of nextRight pointers 
        System.out.println("Following are populated nextRight pointers in \n" + 
      "the tree (-1 is printed if there is no nextRight)"); 
        System.out.println("nextRight of "+ root.data +" is "+ 
        ((root.nextRight != null) ? root.nextRight.data : -1)); 
        System.out.println("nextRight of "+ root.left.data+" is "+ 
        ((root.left.nextRight != null) ? root.left.nextRight.data : -1)); 
        System.out.println("nextRight of "+ root.right.data+" is "+ 
        ((root.right.nextRight != null) ? root.right.nextRight.data : -1)); 
        System.out.println("nextRight of "+  root.left.left.data+" is "+ 
        ((root.left.left.nextRight != null) ? root.left.left.nextRight.data : -1)); 
        System.out.println("nextRight of "+  root.right.right.data+" is "+ 
        ((root.right.right.nextRight != null) ? root.right.right.nextRight.data : -1)); 
    } 
}     
// This code is contributed by Sumit Ghosh 

Python3

#! /usr/bin/env python3 
  
# connect nodes at same level using level order traversal 
import sys 
  
  
class Node: 
    def __init__(self, data): 
        self.data = data 
        self.left = None
        self.right = None
        self.nextRight = None
  
    def __str__(self): 
        return '{}'.format(self.data) 
  
  
def printLevelByLevel(root): 
    # print level by level 
    if root: 
        node = root 
        while node: 
            print('{}'.format(node.data), end=' ') 
            node = node.nextRight 
        print() 
        if root.left: 
            printLevelByLevel(root.left) 
        else: 
            printLevelByLevel(root.right) 
  
  
def inorder(root): 
    if root: 
        inorder(root.left) 
        print(root.data, end=' ') 
        inorder(root.right) 
  
  
def connect(root): 
    # set nextRight of all nodes of a tree 
    queue = [] 
    queue.append(root) 
    # null marker to represent end of current level 
    queue.append(None) 
    # do level order of tree using None markers 
    while queue: 
        p = queue.pop(0) 
        if p: 
            # next element in queue represents 
            # next node at current level 
            p.nextRight = queue[0] 
            # pus left and right children of current node 
            if p.left: 
                queue.append(p.left) 
            if p.right: 
                queue.append(p.right) 
        elif queue: 
            queue.append(None) 
  
  
def main(): 
    """Driver program to test above functions. 
        Constructed binary tree is 
                10 
               /  \ 
             8      2 
            /        \ 
          3            90 
    """
  
    root = Node(10) 
    root.left = Node(8) 
    root.right = Node(2) 
    root.left.left = Node(3) 
    root.right.right = Node(90) 
  
    # Populates nextRight pointer in all nodes 
    connect(root) 
  
    # Let us check the values of nextRight pointers 
    print("Following are populated nextRight pointers in \n"
    "the tree (-1 is printed if there is no nextRight) \n") 
    if(root.nextRight != None): 
        print("nextRight of %d is %d \n" %(root.data,root.nextRight.data)) 
    else: 
        print("nextRight of %d is %d \n" %(root.data,-1)) 
    if(root.left.nextRight != None): 
        print("nextRight of %d is %d \n" %(root.left.data,root.left.nextRight.data)) 
    else: 
        print("nextRight of %d is %d \n" %(root.left.data,-1)) 
    if(root.right.nextRight != None): 
        print("nextRight of %d is %d \n" %(root.right.data,root.right.nextRight.data)) 
    else: 
        print("nextRight of %d is %d \n" %(root.right.data,-1)) 
    if(root.left.left.nextRight != None): 
        print("nextRight of %d is %d \n" %(root.left.left.data,root.left.left.nextRight.data)) 
    else: 
        print("nextRight of %d is %d \n" %(root.left.left.data,-1)) 
    if(root.right.right.nextRight != None): 
        print("nextRight of %d is %d \n" %(root.right.right.data,root.right.right.nextRight.data)) 
    else: 
        print("nextRight of %d is %d \n" %(root.right.right.data,-1)) 
          
    print() 
  
  
if __name__ == "__main__": 
    main() 
  
# This code is contributed by Ram Basnet 

C#

// Connect nodes at same level using level order  
// traversal.  
using System; 
using System.Collections.Generic; 
  
public class Connect_node_same_level  
{  
      
    // Node class  
    class Node 
    {  
        public int data;  
        public Node left, right, nextRight;  
        public Node(int data) 
        {  
            this.data = data;  
            left = null;  
            right = null;  
            nextRight = null;  
        }  
    };  
      
    // Sets nextRight of all nodes of a tree  
    static void connect(Node root)  
    {  
        Queue<Node> q = new Queue<Node>();  
        q.Enqueue(root);  
      
        // null marker to represent end of current level  
        q.Enqueue(null);  
      
        // Do Level order of tree using NULL markers  
        while (q.Count!=0) 
        {  
            Node p = q.Peek();  
            q.Dequeue();  
            if (p != null) 
            {  
      
                // next element in queue represents next  
                // node at current Level  
                p.nextRight = q.Peek();  
      
                // push left and right children of current  
                // node  
                if (p.left != null)  
                    q.Enqueue(p.left);  
                if (p.right != null)  
                    q.Enqueue(p.right);  
            }  
              
            // if queue is not empty, push NULL to mark  
            // nodes at this level are visited  
            else if (q.Count!=0)  
                q.Enqueue(null);  
        }  
    }  
      
    /* Driver code*/
    public static void Main()  
    {  
      
        /* Constructed binary tree is  
                10  
                / \  
            8 2  
            /     \  
        3     90  
        */
        Node root = new Node(10);  
        root.left = new Node(8);  
        root.right = new Node(2);  
        root.left.left = new Node(3);  
        root.right.right = new Node(90);  
      
        // Populates nextRight pointer in all nodes  
        connect(root);  
      
        // Let us check the values of nextRight pointers  
        Console.WriteLine("Following are populated nextRight pointers in \n" +  
    "the tree (-1 is printed if there is no nextRight)");  
        Console.WriteLine("nextRight of "+ root.data +" is "+  
        ((root.nextRight != null) ? root.nextRight.data : -1));  
        Console.WriteLine("nextRight of "+ root.left.data+" is "+  
        ((root.left.nextRight != null) ? root.left.nextRight.data : -1));  
        Console.WriteLine("nextRight of "+ root.right.data+" is "+  
        ((root.right.nextRight != null) ? root.right.nextRight.data : -1));  
        Console.WriteLine("nextRight of "+ root.left.left.data+" is "+  
        ((root.left.left.nextRight != null) ? root.left.left.nextRight.data : -1));  
        Console.WriteLine("nextRight of "+ root.right.right.data+" is "+  
        ((root.right.right.nextRight != null) ? root.right.right.nextRight.data : -1));  
    }  
}  
  
/* This code is contributed by Rajput-Ji*/

Output:

Following are populated nextRight pointers in 
the tree (-1 is printed if there is no nextRight) 
nextRight of 10 is -1 
nextRight of 8 is 2 
nextRight of 2 is -1 
nextRight of 3 is 90 
nextRight of 90 is -1

Time complexity :O(n) where n is the number of nodes

Alternate Implementation:
We can also follow the implementation discussed in Print level order traversal line by line | Set 1. We keep connecting nodes of same level by keeping track of prev visited node of same level.
Implementation : https://ide.geeksforgeeks.org/gV1Oc2

Thanks to Akilan Sengottaiyan for suggesting this alternate implementation.

This article is contributed by Abhishek Rajput. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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My Personal Notes

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python3

C#

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