Connect nodes at same level using constant extra space
Write a function to connect all the adjacent nodes at the same level in a binary tree. The structure of the given Binary Tree node is like following
C++
struct node{ int data; struct node* left; struct node* right; struct node* nextRight;};// This code is contributed by Yash Agarwal(yashagarwal2852002) |
C
struct node { int data; struct node* left; struct node* right; struct node* nextRight;} |
Java
static class node { int data; node left; node right; node nextRight;}// This code is contributed by umadevi9616 |
Python3
class newnode: def __init__(self,data): self.data = data self.left = None self.right = None self.nextRight = None # this code is contributed by shivanisinghss2110 |
C#
public class Node{ public int data; public Node left; public Node right; public Node nextRight;}// this code is contributed by shivanisinghss2110 |
Javascript
class node { constructor() { this.data = 0; this.left = null; this.right = null; this.nextRight = null; }}// This code is contributed by importantly. |
Initially, all the nextRight pointers point to garbage values. Your function should set these pointers to point next right for each node. You can use only constant extra space.
Example:
Input Tree
A
/ \
B C
/ \ \
D E F
Output Tree
A--->NULL
/ \
B-->C-->NULL
/ \ \
D-->E-->F-->NULLWe discussed two different approaches to do it in the previous post. The auxiliary space required in both of those approaches is not constant. Also, method 2 discussed there only works for a complete Binary Tree.
In this post, we will first modify method 2 to make it work for all kinds of trees. After that, we will remove recursion from this method so that the extra space becomes constant.
A Recursive Solution
In method 2 of previous post, we traversed the nodes in pre-order fashion. Instead of traversing in Pre Order fashion (root, left, right), if we traverse the nextRight node before the left and right children (root, nextRight, left), then we can make sure that all nodes at level i have the nextRight set, before the level i+1 nodes. Let us consider the following example (same example as previous post). Method 2 fails for right child of node 4. In this method, we make sure that all nodes at the 4’s level (level 2) have nextRight set, before we try to set the nextRight of 9. So when we set the nextRight of 9, we search for a nonleaf node on right side of node 4 (getNextRight() does this for us).
1 -------------- Level 0
/ \
2 3 -------------- Level 1
/ \ / \
4 5 6 7 -------------- Level 2
/ \ / \
8 9 10 11 -------------- Level 3C++
// THIS CODE IS CONTRIBUTED BY YASH AGARWAL#include<bits/stdc++.h>using namespace std; class Node { public: int data; Node *left, *right, *nextRight; Node(int item){ data = item; left = right = nextRight = nullptr; }}; class BinaryTree{ public: Node *root; BinaryTree(){ root = nullptr; } void connectRecur(Node *p) { // Base case if (p == NULL) return; /* Before setting nextRight of left and right children, set nextRight of children of other nodes at same level (because we can access children of other nodes using p's nextRight only) */ if (p->nextRight) connectRecur(p->nextRight); /* Set the nextRight pointer for p's left child */ if (p->left) { if (p->right) { p->left->nextRight = p->right; p->right->nextRight = getNextRight(p); }else{ p->left->nextRight = getNextRight(p); } // Recursively call for next level nodes.Note that we call only // left child. The call for left child will call for right child connectRecur(p->left); } /* If left child is NULL then first node of next level will either be p->right or getNextRight(p) */ else if (p->right) { p->right->nextRight = getNextRight(p); connectRecur(p->right); } else connectRecur(getNextRight(p)); } // This function returns the leftmost child of nodes at the same level // as p. This function is used to getNExt right of p's right childIf // right child of p is NULL then this can also be used for the left child Node *getNextRight(Node *p) { Node *temp = p->nextRight; /* Traverse nodes at p's level and find and return the first node's first child */ while (temp) { if (temp->left) return temp->left; if (temp->right) return temp->right; temp = temp->nextRight; } // If all the nodes at p's level are leaf nodes then return NULL return nullptr; } void connect() { // Set the nextRight for root root->nextRight = nullptr; // Set the next right for rest of the // nodes (other than root) connectRecur(root); }}; int main(){ BinaryTree tree; tree.root = new Node(10); tree.root->left = new Node(8); tree.root->right = new Node(2); tree.root->left->left = new Node(3); tree.root->right->right = new Node(90); // Populates nextRight pointer in all nodes tree.connectRecur(tree.root); // Let us check the values of nextRight pointers int a = tree.root->nextRight != NULL ? tree.root->nextRight->data : -1; int b = tree.root->left->nextRight != NULL ? tree.root->left->nextRight->data : -1; int c = tree.root->right->nextRight != NULL ? tree.root->right->nextRight->data : -1; int d = tree.root->left->left->nextRight != NULL ? tree.root->left->left->nextRight->data : -1; int e = tree.root->right->right->nextRight != NULL ? tree.root->right->right->nextRight->data : -1; // Now lets print the values cout << "Following are populated nextRight pointers in the tree(-1 is printed if there is no nextRight) \n" << "nextRight of " << tree.root->data << " is " << a << "\n" << "nextRight of " << tree.root->left->data << " is " << b << "\n" << "nextRight of " << tree.root->right->data << " is " << c << "\n" << "nextRight of " << tree.root->left->left->data << " is " << d << "\n" << "nextRight of " << tree.root->right->right->data << " is " << e << "\n"; return 0;} |
C
// Recursive C program to connect nodes at same level// using constant extra spacevoid connectRecur(struct node* p);struct node *getNextRight(struct node *p);// Sets the nextRight of root and calls// connectRecur() for other nodesvoid connect (struct node *p){ // Set the nextRight for root p->nextRight = NULL; // Set the next right for rest of the // nodes (other than root) connectRecur(p);}/* Set next right of all descendants of p. This function makes sure that nextRight of nodes at level i is set before level i+1 nodes. */void connectRecur(struct node* p){ // Base case if (!p) return; /* Set the nextRight pointer for p's left child */ if (p->left) { if (p->right) { p->left->nextRight = p->right; p->right->nextRight = getNextRight(p); connectRecur(p->right); } else p->left->nextRight = getNextRight(p); /* Recursively call for next level nodes. Note that we call only for left child. The call for left child will call for right child */ connectRecur(p->left); } /* If left child is NULL then first node of next level will either be p->right or getNextRight(p) */ else if (p->right) { p->right->nextRight = getNextRight(p); connectRecur(p->right); }}/* This function returns the leftmost child of nodes at the same level as p. This function is used to getNExt right of p's right child If right child of p is NULL then this can also be used for the left child */struct node *getNextRight(struct node *p){ struct node *temp = p->nextRight; /* Traverse nodes at p's level and find and return the first node's first child */ while(temp != NULL) { if(temp->left != NULL) return temp->left; if(temp->right != NULL) return temp->right; temp = temp->nextRight; } // If all the nodes at p's level are leaf // nodes then return NULL return NULL;} |
Java
// Recursive Java program to connect nodes at same level// using constant extra space // A binary tree nodeclass Node{ int data; Node left, right, nextRight; Node(int item) { data = item; left = right = nextRight = null; }} class BinaryTree{ Node root; /* Set next right of all descendants of p. This function makes sure that nextRight of nodes at level i is set before level i+1 nodes. */ void connectRecur(Node p) { // Base case if (p == null) return; /* Before setting nextRight of left and right children, set nextRight of children of other nodes at same level (because we can access children of other nodes using p's nextRight only) */ if (p.nextRight != null) connectRecur(p.nextRight); /* Set the nextRight pointer for p's left child */ if (p.left != null) { if (p.right != null) { p.left.nextRight = p.right; p.right.nextRight = getNextRight(p); } else p.left.nextRight = getNextRight(p); /* Recursively call for next level nodes. Note that we call only for left child. The call for left child will call for right child */ connectRecur(p.left); } /* If left child is NULL then first node of next level will either be p->right or getNextRight(p) */ else if (p.right != null) { p.right.nextRight = getNextRight(p); connectRecur(p.right); } else connectRecur(getNextRight(p)); } /* This function returns the leftmost child of nodes at the same level as p. This function is used to getNExt right of p's right child If right child of p is NULL then this can also be used for the left child */ Node getNextRight(Node p) { Node temp = p.nextRight; /* Traverse nodes at p's level and find and return the first node's first child */ while (temp != null) { if (temp.left != null) return temp.left; if (temp.right != null) return temp.right; temp = temp.nextRight; } // If all the nodes at p's level are leaf nodes then return NULL return null; } /* Driver program to test the above functions */ public static void main(String args[]) { BinaryTree tree = new BinaryTree(); tree.root = new Node(10); tree.root.left = new Node(8); tree.root.right = new Node(2); tree.root.left.left = new Node(3); tree.root.right.right = new Node(90); // Populates nextRight pointer in all nodes tree.connectRecur(tree.root); // Let us check the values of nextRight pointers int a = tree.root.nextRight != null ? tree.root.nextRight.data : -1; int b = tree.root.left.nextRight != null ? tree.root.left.nextRight.data : -1; int c = tree.root.right.nextRight != null ? tree.root.right.nextRight.data : -1; int d = tree.root.left.left.nextRight != null ? tree.root.left.left.nextRight.data : -1; int e = tree.root.right.right.nextRight != null ? tree.root.right.right.nextRight.data : -1; // Now lets print the values System.out.println("Following are populated nextRight pointers in " + " the tree(-1 is printed if there is no nextRight)"); System.out.println("nextRight of " + tree.root.data + " is " + a); System.out.println("nextRight of " + tree.root.left.data + " is " + b); System.out.println("nextRight of " + tree.root.right.data + " is " + c); System.out.println("nextRight of " + tree.root.left.left.data + " is " + d); System.out.println("nextRight of " + tree.root.right.right.data + " is " + e); }}// This code has been contributed by Mayank Jaiswal |
Python3
# Recursive Python program to connect nodes# at same level using constant extra space# Helper class that allocates a new node# with the given data and None left and# right pointers.class newnode: def __init__(self,data): self.data = data self.left = None self.right = None self.nextRight = None # Set next right of all descendants of p. This function makes sure that# nextRight of nodes at level i is set before level i+1 nodes. */def connectRecur( p) : # Base case if p == None : return # Before setting nextRight of left and right children, set nextRight # of children of other nodes at same level (because we can access # children of other nodes using p's nextRight only) */ if p.nextRight != None : connectRecur(p.nextRight) # Set the nextRight pointer for p's left child */ if p.left != None : if p.right != None : p.left.nextRight = p.right p.right.nextRight = getNextRight(p) else : p.left.nextRight = getNextRight(p) # Recursively call for next level nodes. Note that we call only # for left child. The call for left child will call for right child */ connectRecur(p.left) # If left child is NULL then first node of next level will either be # p->right or getNextRight(p) */ elif p.right != None : p.right.nextRight = getNextRight(p) connectRecur(p.right) else : connectRecur(getNextRight(p)) # This function returns the leftmost child of# nodes at the same level as p. This function# is used to getNExt right of p's right child# If right child of is None then this can also# be used for the left childdef getNextRight(p): temp = p.nextRight # Traverse nodes at p's level and find # and return the first node's first child while (temp != None): if (temp.left != None): return temp.left if (temp.right != None): return temp.right temp = temp.nextRight # If all the nodes at p's level are # leaf nodes then return None return None# Driver Codeif __name__ == '__main__': # Constructed binary tree is # 10 # / \ # 8 2 # / \ # 3 90 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 connectRecur(root) # Let us check the values of nextRight pointers print("Following are populated nextRight " "pointers in the tree (-1 is printed " "if there is no nextRight) \n") print("nextRight of", root.data, "is", end = " ") if root.nextRight: print(root.nextRight.data) else: print(-1) print("nextRight of", root.left.data, "is", end = " ") if root.left.nextRight: print(root.left.nextRight.data) else: print(-1) print("nextRight of", root.right.data, "is", end = " ") if root.right.nextRight: print(root.right.nextRight.data) else: print(-1) print("nextRight of", root.left.left.data, "is", end = " ") if root.left.left.nextRight: print(root.left.left.nextRight.data) else: print(-1) print("nextRight of", root.right.right.data, "is", end = " ") if root.right.right.nextRight: print(root.right.right.nextRight.data) else: print(-1)# This code is contributed by jana_sayantan. |
C#
using System;// Recursive C# program to connect nodes at same level// using constant extra space// A binary tree nodepublic class Node{ public int data; public Node left, right, nextRight; public Node(int item) { data = item; left = right = nextRight = null; }}public class BinaryTree{ public Node root; /* Set next right of all descendants of p. This function makes sure that nextRight of nodes at level i is set before level i+1 nodes. */ public virtual void connectRecur(Node p) { // Base case if (p == null) { return; } /* Before setting nextRight of left and right children, set nextRight of children of other nodes at same level (because we can access children of other nodes using p's nextRight only) */ if (p.nextRight != null) { connectRecur(p.nextRight); } /* Set the nextRight pointer for p's left child */ if (p.left != null) { if (p.right != null) { p.left.nextRight = p.right; p.right.nextRight = getNextRight(p); } else { p.left.nextRight = getNextRight(p); } /* Recursively call for next level nodes. Note that we call only for left child. The call for left child will call for right child */ connectRecur(p.left); } /* If left child is NULL then first node of next level will either be p->right or getNextRight(p) */ else if (p.right != null) { p.right.nextRight = getNextRight(p); connectRecur(p.right); } else { connectRecur(getNextRight(p)); } } /* This function returns the leftmost child of nodes at the same level as p. This function is used to getNExt right of p's right child If right child of p is NULL then this can also be used for the left child */ public virtual Node getNextRight(Node p) { Node temp = p.nextRight; /* Traverse nodes at p's level and find and return the first node's first child */ while (temp != null) { if (temp.left != null) { return temp.left; } if (temp.right != null) { return temp.right; } temp = temp.nextRight; } // If all the nodes at p's level are leaf nodes then return NULL return null; } /* Driver program to test the above functions */ public static void Main(string[] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(10); tree.root.left = new Node(8); tree.root.right = new Node(2); tree.root.left.left = new Node(3); tree.root.right.right = new Node(90); // Populates nextRight pointer in all nodes tree.connectRecur(tree.root); // Let us check the values of nextRight pointers int a = tree.root.nextRight != null ? tree.root.nextRight.data : -1; int b = tree.root.left.nextRight != null ? tree.root.left.nextRight.data : -1; int c = tree.root.right.nextRight != null ? tree.root.right.nextRight.data : -1; int d = tree.root.left.left.nextRight != null ? tree.root.left.left.nextRight.data : -1; int e = tree.root.right.right.nextRight != null ? tree.root.right.right.nextRight.data : -1; // Now lets print the values Console.WriteLine("Following are populated nextRight pointers in the tree(-1 is printed if there is no nextRight)"); Console.WriteLine("nextRight of " + tree.root.data + " is " + a); Console.WriteLine("nextRight of " + tree.root.left.data + " is " + b); Console.WriteLine("nextRight of " + tree.root.right.data + " is " + c); Console.WriteLine("nextRight of " + tree.root.left.left.data + " is " + d); Console.WriteLine("nextRight of " + tree.root.right.right.data + " is " + e); }} // This code is contributed by Shrikant13 |
Javascript
<script>// Recursive javascript program to connect nodes at same level// using constant extra space // A binary tree nodeclass Node { constructor(val) { this.data = val; this.left = null; this.right = null; this.nextRight = null; } } var root; /* Set next right of all descendants of p. This function makes sure that nextRight of nodes ar level i is set before level i+1 nodes. */ function connectRecur(p) { // Base case if (p == null) return; /* Before setting nextRight of left and right children, set nextRight of children of other nodes at same level (because we can access children of other nodes using p's nextRight only) */ if (p.nextRight != null) connectRecur(p.nextRight); /* Set the nextRight pointer for p's left child */ if (p.left != null) { if (p.right != null) { p.left.nextRight = p.right; p.right.nextRight = getNextRight(p); } else p.left.nextRight = getNextRight(p); /* Recursively call for next level nodes. Note that we call only for left child. The call for left child will call for right child */ connectRecur(p.left); } /* If left child is NULL then first node of next level will either be p->right or getNextRight(p) */ else if (p.right != null) { p.right.nextRight = getNextRight(p); connectRecur(p.right); } else connectRecur(getNextRight(p)); } /* This function returns the leftmost child of nodes at the same level as p. This function is used to getNExt right of p's right child If right child of p is NULL then this can also be used for the left child */ function getNextRight(p) { var temp = p.nextRight; /* Traverse nodes at p's level and find and return the first node's first child */ while (temp != null) { if (temp.left != null) return temp.left; if (temp.right != null) return temp.right; temp = temp.nextRight; } // If all the nodes at p's level are leaf nodes then return NULL return null; } /* Driver program to test the above functions */ var 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 connectRecur(root); // Let us check the values of nextRight pointers var a = root.nextRight != null ? root.nextRight.data : -1; var b = root.left.nextRight != null ? root.left.nextRight.data : -1; var c = root.right.nextRight != null ? root.right.nextRight.data : -1; var d = root.left.left.nextRight != null ? root.left.left.nextRight.data : -1; var e = root.right.right.nextRight != null ? root.right.right.nextRight.data : -1; // Now lets print the values document.write("Following are populated nextRight pointers in " + " the tree(-1 is printed if there is no nextRight)"); document.write("<br/>nextRight of " + root.data + " is " + a); document.write("<br/>nextRight of " + root.left.data + " is " + b); document.write("<br/>nextRight of " + root.right.data + " is " + c); document.write("<br/>nextRight of " + root.left.left.data + " is " + d); document.write("<br/>nextRight of " + root.right.right.data + " is " + e);// This code is contributed by umadevi9616</script> |
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)
Auxiliary Space: O(1)
An Iterative Solution :
The recursive approach discussed above can be easily converted to iterative. In the iterative version, we use nested loop. The outer loop, goes through all the levels and the inner loop goes through all the nodes at every level. This solution uses constant space.
C++
// Iterative CPP program to connect// nodes at same level using// constant extra space#include<bits/stdc++.h>#include<bits/stdc++.h>using namespace std;class node{ public: int data; node* left; node* right; node *nextRight; /* Constructor that allocates a new node with the given data and NULL left and right pointers. */ node(int data) { this->data = data; this->left = NULL; this->right = NULL; this->nextRight = NULL; }};/* This function returns the leftmostchild of nodes at the same level as p.This function is used to getNExt rightof p's right child If right child ofis NULL then this can also be used forthe left child */node *getNextRight(node *p){ node *temp = p->nextRight; /* Traverse nodes at p's level and find and return the first node's first child */ while (temp != NULL) { if (temp->left != NULL) return temp->left; if (temp->right != NULL) return temp->right; temp = temp->nextRight; } // If all the nodes at p's level // are leaf nodes then return NULL return NULL;}/* Sets nextRight of all nodesof a tree with root as p */void connectRecur(node* p){ node *temp; if (!p) return; // Set nextRight for root p->nextRight = NULL; // set nextRight of all levels one by one while (p != NULL) { node *q = p; /* Connect all children nodes of p and children nodes of all other nodes at same level as p */ while (q != NULL) { // Set the nextRight pointer // for p's left child if (q->left) { // If q has right child, then // right child is nextRight of // p and we also need to set // nextRight of right child if (q->right) q->left->nextRight = q->right; else q->left->nextRight = getNextRight(q); } if (q->right) q->right->nextRight = getNextRight(q); // Set nextRight for other // nodes in pre order fashion q = q->nextRight; } // start from the first // node of next level if (p->left) p = p->left; else if (p->right) p = p->right; else p = getNextRight(p); }}/* Driver code*/int 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 connectRecur(root); // Let us check the values of nextRight pointers cout << "Following are populated nextRight pointers in the tree" " (-1 is printed if there is no nextRight) \n"; cout << "nextRight of " << root->data << " is " << (root->nextRight? root->nextRight->data: -1) <<endl; cout << "nextRight of " << root->left->data << " is " << (root->left->nextRight? root->left->nextRight->data: -1) << endl; cout << "nextRight of " << root->right->data << " is " << (root->right->nextRight? root->right->nextRight->data: -1) << endl; cout << "nextRight of " << root->left->left->data<< " is " << (root->left->left->nextRight? root->left->left->nextRight->data: -1) << endl; cout << "nextRight of " << root->right->right->data << " is " << (root->right->right->nextRight? root->right->right->nextRight->data: -1) << endl; return 0;}// This code is contributed by rathbhupendra |
C
// Iterative C program to connect nodes at same level// using constant extra space#include <stdio.h>#include <stdlib.h>struct node{ int data; struct node *left; struct node *right; struct node *nextRight;};/* This function returns the leftmost child of nodes at the same level as p. This function is used to getNExt right of p's right child If right child of is NULL then this can also be used for the left child */struct node *getNextRight(struct node *p){ struct node *temp = p->nextRight; /* Traverse nodes at p's level and find and return the first node's first child */ while (temp != NULL) { if (temp->left != NULL) return temp->left; if (temp->right != NULL) return temp->right; temp = temp->nextRight; } // If all the nodes at p's level are leaf nodes then return NULL return NULL;}/* Sets nextRight of all nodes of a tree with root as p */void connect(struct node* p){ struct node *temp; if (!p) return; // Set nextRight for root p->nextRight = NULL; // set nextRight of all levels one by one while (p != NULL) { struct node *q = p; /* Connect all children nodes of p and children nodes of all other nodes at same level as p */ while (q != NULL) { // Set the nextRight pointer for p's left child if (q->left) { // If q has right child, then right child is nextRight of // p and we also need to set nextRight of right child if (q->right) q->left->nextRight = q->right; else q->left->nextRight = getNextRight(q); } if (q->right) q->right->nextRight = getNextRight(q); // Set nextRight for other nodes in pre order fashion q = q->nextRight; } // start from the first node of next level if (p->left) p = p->left; else if (p->right) p = p->right; else p = getNextRight(p); }}/* 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 = NULL; node->right = NULL; 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 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); getchar(); return 0;} |
Java
// Iterative Java program to connect nodes at same level// using constant extra space // A binary tree nodeclass Node{ int data; Node left, right, nextRight; Node(int item) { data = item; left = right = nextRight = null; }} class BinaryTree{ Node root; /* This function returns the leftmost child of nodes at the same level as p. This function is used to getNExt right of p's right child If right child of is NULL then this can also be used for the left child */ Node getNextRight(Node p) { Node temp = p.nextRight; /* Traverse nodes at p's level and find and return the first node's first child */ while (temp != null) { if (temp.left != null) return temp.left; if (temp.right != null) return temp.right; temp = temp.nextRight; } // If all the nodes at p's level are leaf nodes then return NULL return null; } /* Sets nextRight of all nodes of a tree with root as p */ void connect(Node p) { Node temp = null; if (p == null) return; // Set nextRight for root p.nextRight = null; // set nextRight of all levels one by one while (p != null) { Node q = p; /* Connect all children nodes of p and children nodes of all other nodes at same level as p */ while (q != null) { // Set the nextRight pointer for p's left child if (q.left != null) { // If q has right child, then right child is nextRight of // p and we also need to set nextRight of right child if (q.right != null) q.left.nextRight = q.right; else q.left.nextRight = getNextRight(q); } if (q.right != null) q.right.nextRight = getNextRight(q); // Set nextRight for other nodes in pre order fashion q = q.nextRight; } // start from the first node of next level if (p.left != null) p = p.left; else if (p.right != null) p = p.right; else p = getNextRight(p); } } /* Driver program to test above functions */ public static void main(String args[]) { /* Constructed binary tree is 10 / \ 8 2 / \ 3 90 */ BinaryTree tree = new BinaryTree(); tree.root = new Node(10); tree.root.left = new Node(8); tree.root.right = new Node(2); tree.root.left.left = new Node(3); tree.root.right.right = new Node(90); // Populates nextRight pointer in all nodes tree.connect(tree.root); // Let us check the values of nextRight pointers int a = tree.root.nextRight != null ? tree.root.nextRight.data : -1; int b = tree.root.left.nextRight != null ? tree.root.left.nextRight.data : -1; int c = tree.root.right.nextRight != null ? tree.root.right.nextRight.data : -1; int d = tree.root.left.left.nextRight != null ? tree.root.left.left.nextRight.data : -1; int e = tree.root.right.right.nextRight != null ? tree.root.right.right.nextRight.data : -1; // Now lets print the values System.out.println("Following are populated nextRight pointers in " + " the tree(-1 is printed if there is no nextRight)"); System.out.println("nextRight of " + tree.root.data + " is " + a); System.out.println("nextRight of " + tree.root.left.data + " is " + b); System.out.println("nextRight of " + tree.root.right.data + " is " + c); System.out.println("nextRight of " + tree.root.left.left.data + " is " + d); System.out.println("nextRight of " + tree.root.right.right.data + " is " + e); }}// This code has been contributed by Mayank Jaiswal |
Python3
# Iterative Python program to connect nodes# at same level using constant extra space# Helper class that allocates a new node# with the given data and None left and# right pointers.class newnode: def __init__(self,data): self.data = data self.left = None self.right = None self.nextRight = None # This function returns the leftmost child of # nodes at the same level as p. This function# is used to getNExt right of p's right child# If right child of is None then this can also# be used for the left childdef getNextRight(p): temp = p.nextRight # Traverse nodes at p's level and find # and return the first node's first child while (temp != None): if (temp.left != None): return temp.left if (temp.right != None): return temp.right temp = temp.nextRight # If all the nodes at p's level are # leaf nodes then return None return None# Sets nextRight of all nodes of a tree# with root as pdef connect(p): temp = None if (not p): return # Set nextRight for root p.nextRight = None # set nextRight of all levels one by one while (p != None): q = p # Connect all children nodes of p and # children nodes of all other nodes # at same level as p while (q != None): # Set the nextRight pointer for # p's left child if (q.left): # If q has right child, then right # child is nextRight of p and we # also need to set nextRight of # right child if (q.right): q.left.nextRight = q.right else: q.left.nextRight = getNextRight(q) if (q.right): q.right.nextRight = getNextRight(q) # Set nextRight for other nodes in # pre order fashion q = q.nextRight # start from the first node # of next level if (p.left): p = p.left elif (p.right): p = p.right else: p = getNextRight(p)# Driver Codeif __name__ == '__main__': # Constructed binary tree is # 10 # / \ # 8 2 # / \ # 3 90 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 print("Following are populated nextRight " "pointers in the tree (-1 is printed " "if there is no nextRight) \n") print("nextRight of", root.data, "is", end = " ") if root.nextRight: print(root.nextRight.data) else: print(-1) print("nextRight of", root.left.data, "is", end = " ") if root.left.nextRight: print(root.left.nextRight.data) else: print(-1) print("nextRight of", root.right.data, "is", end = " ") if root.right.nextRight: print(root.right.nextRight.data) else: print(-1) print("nextRight of", root.left.left.data, "is", end = " ") if root.left.left.nextRight: print(root.left.left.nextRight.data) else: print(-1) print("nextRight of", root.right.right.data, "is", end = " ") if root.right.right.nextRight: print(root.right.right.nextRight.data) else: print(-1)# This code is contributed by PranchalK |
C#
using System;// Iterative c# program to connect nodes at same level// using constant extra space// A binary tree nodepublic class Node{ public int data; public Node left, right, nextRight; public Node(int item) { data = item; left = right = nextRight = null; }}public class BinaryTree{ public Node root; /* This function returns the leftmost child of nodes at the same level as p. This function is used to getNExt right of p's right child If right child of is NULL then this can also be used for the left child */ public virtual Node getNextRight(Node p) { Node temp = p.nextRight; /* Traverse nodes at p's level and find and return the first node's first child */ while (temp != null) { if (temp.left != null) { return temp.left; } if (temp.right != null) { return temp.right; } temp = temp.nextRight; } // If all the nodes at p's level are leaf nodes then return NULL return null; } /* Sets nextRight of all nodes of a tree with root as p */ public virtual void connect(Node p) { Node temp = null; if (p == null) { return; } // Set nextRight for root p.nextRight = null; // set nextRight of all levels one by one while (p != null) { Node q = p; /* Connect all children nodes of p and children nodes of all other nodes at same level as p */ while (q != null) { // Set the nextRight pointer for p's left child if (q.left != null) { // If q has right child, then right child is nextRight of // p and we also need to set nextRight of right child if (q.right != null) { q.left.nextRight = q.right; } else { q.left.nextRight = getNextRight(q); } } if (q.right != null) { q.right.nextRight = getNextRight(q); } // Set nextRight for other nodes in pre order fashion q = q.nextRight; } // start from the first node of next level if (p.left != null) { p = p.left; } else if (p.right != null) { p = p.right; } else { p = getNextRight(p); } } } /* Driver program to test above functions */ public static void Main(string[] args) { /* Constructed binary tree is 10 / \ 8 2 / \ 3 90 */ BinaryTree tree = new BinaryTree(); tree.root = new Node(10); tree.root.left = new Node(8); tree.root.right = new Node(2); tree.root.left.left = new Node(3); tree.root.right.right = new Node(90); // Populates nextRight pointer in all nodes tree.connect(tree.root); // Let us check the values of nextRight pointers int a = tree.root.nextRight != null ? tree.root.nextRight.data : -1; int b = tree.root.left.nextRight != null ? tree.root.left.nextRight.data : -1; int c = tree.root.right.nextRight != null ? tree.root.right.nextRight.data : -1; int d = tree.root.left.left.nextRight != null ? tree.root.left.left.nextRight.data : -1; int e = tree.root.right.right.nextRight != null ? tree.root.right.right.nextRight.data : -1; // Now lets print the values Console.WriteLine("Following are populated nextRight pointers in " + " the tree(-1 is printed if there is no nextRight)"); Console.WriteLine("nextRight of " + tree.root.data + " is " + a); Console.WriteLine("nextRight of " + tree.root.left.data + " is " + b); Console.WriteLine("nextRight of " + tree.root.right.data + " is " + c); Console.WriteLine("nextRight of " + tree.root.left.left.data + " is " + d); Console.WriteLine("nextRight of " + tree.root.right.right.data + " is " + e); }} // This code is contributed by Shrikant13 |
Javascript
<script> // Iterative Javascript program to connect nodes at same level // using constant extra space class Node { constructor(item) { this.data = item; this.left = null; this.right = null; this.nextRight = null; } } let root; /* This function returns the leftmost child of nodes at the same level as p. This function is used to getNExt right of p's right child If right child of is NULL then this can also be used for the left child */ function getNextRight(p) { let temp = p.nextRight; /* Traverse nodes at p's level and find and return the first node's first child */ while (temp != null) { if (temp.left != null) return temp.left; if (temp.right != null) return temp.right; temp = temp.nextRight; } // If all the nodes at p's level are // leaf nodes then return NULL return null; } /* Sets nextRight of all nodes of a tree with root as p */ function connect(p) { let temp = null; if (p == null) return; // Set nextRight for root p.nextRight = null; // set nextRight of all levels one by one while (p != null) { let q = p; /* Connect all children nodes of p and children nodes of all other nodes at same level as p */ while (q != null) { // Set the nextRight pointer for p's left child if (q.left != null) { // If q has right child, then // right child is nextRight of // p and we also need to set // nextRight of right child if (q.right != null) q.left.nextRight = q.right; else q.left.nextRight = getNextRight(q); } if (q.right != null) q.right.nextRight = getNextRight(q); // Set nextRight for other nodes // in pre order fashion q = q.nextRight; } // start from the first node of next level if (p.left != null) p = p.left; else if (p.right != null) p = p.right; else p = getNextRight(p); } } /* Constructed binary tree is 10 / \ 8 2 / \ 3 90 */ 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 let a = root.nextRight != null ? root.nextRight.data : -1; let b = root.left.nextRight != null ? root.left.nextRight.data : -1; let c = root.right.nextRight != null ? root.right.nextRight.data : -1; let d = root.left.left.nextRight != null ? root.left.left.nextRight.data : -1; let e = root.right.right.nextRight != null ? root.right.right.nextRight.data : -1; // Now lets print the values document.write("Following are populated nextRight pointers in " + " the tree(-1 is printed if there is no nextRight)" + "</br>"); document.write("nextRight of " + root.data + " is " + a + "</br>"); document.write("nextRight of " + root.left.data + " is " + b + "</br>"); document.write("nextRight of " + root.right.data + " is " + c + "</br>"); document.write("nextRight of " + root.left.left.data + " is " + d + "</br>"); document.write("nextRight of " + root.right.right.data + " is " + e + "</br>"); </script> |
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)
Auxiliary Space: O(1)
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