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
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.
Implementation:
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) else if 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*/ |
Javascript
<script> // Connect nodes at same level using level order traversal. // A Binary Tree Node class Node { constructor(data, nextRight) { this .left = null ; this .right = null ; this .data = data; this .nextRight = nextRight; } } // Sets nextRight of all nodes of a tree function connect(root) { let 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.length > 0) { let p = q[0]; q.shift(); if (p != null ) { // next element in queue represents next // node at current Level p.nextRight = q[0]; // push left and right children of current // node if (p.left != null ) q.push(p.left); if (p.right != null ) q.push(p.right); } // if queue is not empty, push NULL to mark // nodes at this level are visited else if (q.length > 0) q.push( null ); } } /* Constructed binary tree is 10 / \ 8 2 / \ 3 90 */ let 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 document.write( "Following are populated nextRight pointers in " + "</br>" + "the tree (-1 is printed if there is no nextRight)" + "</br>" ); document.write( "nextRight of " + root.data + " is " + ((root.nextRight != null ) ? root.nextRight.data : -1) + "</br>" ); document.write( "nextRight of " + root.left.data+ " is " + ((root.left.nextRight != null ) ? root.left.nextRight.data : -1) + "</br>" ); document.write( "nextRight of " + root.right.data+ " is " + ((root.right.nextRight != null ) ? root.right.nextRight.data : -1) + "</br>" ); document.write( "nextRight of " + root.left.left.data+ " is " + ((root.left.left.nextRight != null ) ? root.left.left.nextRight.data : -1) + "</br>" ); document.write( "nextRight of " + root.right.right.data+ " is " + ((root.right.right.nextRight != null ) ? root.right.right.nextRight.data : -1) + "</br>" ); // This code is contributed by divyesh072019. </script> |
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
Auxiliary Space: O(n) for queue
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 previous 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 write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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