Populate Inorder Successor for all nodes
Given a Binary Tree where each node has the following structure, write a function to populate the next pointer for all nodes. The next pointer for every node should be set to point to inorder successor.
C++
class node
{
public:
int data;
node *left;
node *right;
node *next;
};
// This code is contributed
// by Shubham Singh
C
struct node
{
int data;
struct node* left;
struct node* right;
struct node* next;
}
Java
// A binary tree node
class Node
{
int data;
Node left, right, next;
Node(int item)
{
data = item;
left = right = next = null;
}
}
// This code is contributed by SUBHAMSINGH10.
Python3
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
self.next = None
# This code is contributed by Shubham Singh
C#
class Node
{
public int data;
public Node left, right, next;
public Node(int item)
{
data = item;
left = right = next = null;
}
}
Node root;
// This code is contributed
// by Shubham Singh
Javascript
<script>
class Node
{
constructor(x) {
this.data = x;
this.left = null;
this.right = null;
}
}
// This code is contributed by Shubham Singh
</script>
Initially, all next pointers have NULL values. Your function should fill these next pointers so that they point to inorder successor.
Solution (Use Reverse Inorder Traversal)
Traverse the given tree in reverse inorder traversal and keep track of previously visited node. When a node is being visited, assign a previously visited node as next.
C++
// C++ program to populate inorder
// traversal of all nodes
#include<bits/stdc++.h>
using namespace std;
class node
{
public:
int data;
node *left;
node *right;
node *next;
};
/* Set next of p and all descendants of p
by traversing them in reverse Inorder */
void populateNext(node* p)
{
// The first visited node will be the
// rightmost node next of the rightmost
// node will be NULL
static node *next = NULL;
if (p)
{
// First set the next pointer
// in right subtree
populateNext(p->right);
// Set the next as previously visited
// node in reverse Inorder
p->next = next;
// Change the prev for subsequent node
next = p;
// Finally, set the next pointer in
// left subtree
populateNext(p->left);
}
}
/* UTILITY FUNCTIONS */
/* Helper function that allocates a new
node with the given data and NULL left
and right pointers. */
node* newnode(int data)
{
node* Node = new node();
Node->data = data;
Node->left = NULL;
Node->right = NULL;
Node->next = NULL;
return(Node);
}
// Driver Code
int main()
{
/* Constructed binary tree is
10
/ \
8 12
/
3
*/
node *root = newnode(10);
root->left = newnode(8);
root->right = newnode(12);
root->left->left = newnode(3);
// Populates nextRight pointer in all nodes
populateNext(root);
// Let us see the populated values
node *ptr = root->left->left;
while(ptr)
{
// -1 is printed if there is no successor
cout << "Next of " << ptr->data << " is "
<< (ptr->next? ptr->next->data: -1)
<< endl;
ptr = ptr->next;
}
return 0;
}
// This code is contributed by rathbhupendra
Java
// Java program to populate inorder traversal of all nodes
// A binary tree node
class Node
{
int data;
Node left, right, next;
Node(int item)
{
data = item;
left = right = next = null;
}
}
class BinaryTree
{
Node root;
static Node next = null;
/* Set next of p and all descendants of p by traversing them in
reverse Inorder */
void populateNext(Node node)
{
// The first visited node will be the rightmost node
// next of the rightmost node will be NULL
if (node != null)
{
// First set the next pointer in right subtree
populateNext(node.right);
// Set the next as previously visited node in reverse Inorder
node.next = next;
// Change the prev for subsequent node
next = node;
// Finally, set the next pointer in left subtree
populateNext(node.left);
}
}
/* Driver program to test above functions*/
public static void main(String args[])
{
/* Constructed binary tree is
10
/ \
8 12
/
3 */
BinaryTree tree = new BinaryTree();
tree.root = new Node(10);
tree.root.left = new Node(8);
tree.root.right = new Node(12);
tree.root.left.left = new Node(3);
// Populates nextRight pointer in all nodes
tree.populateNext(tree.root);
// Let us see the populated values
Node ptr = tree.root.left.left;
while (ptr != null)
{
// -1 is printed if there is no successor
int print = ptr.next != null ? ptr.next.data : -1;
System.out.println("Next of " + ptr.data + " is: " + print);
ptr = ptr.next;
}
}
}
// This code has been contributed by Mayank Jaiswal
Python3
# Python3 program to populate
# inorder traversal of all nodes
# Tree node
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
self.next = None
# The first visited node will be
# the rightmost node next of the
# rightmost node will be None
next = None
# Set next of p and all descendants of p
# by traversing them in reverse Inorder
def populateNext(p):
global next
if (p != None):
# First set the next pointer
# in right subtree
populateNext(p.right)
# Set the next as previously visited node
# in reverse Inorder
p.next = next
# Change the prev for subsequent node
next = p
# Finally, set the next pointer
# in left subtree
populateNext(p.left)
# UTILITY FUNCTIONS
# Helper function that allocates
# a new node with the given data
# and None left and right pointers.
def newnode(data):
node = Node(0)
node.data = data
node.left = None
node.right = None
node.next = None
return(node)
# Driver Code
# Constructed binary tree is
# 10
# / \
# 8 12
# /
# 3
root = newnode(10)
root.left = newnode(8)
root.right = newnode(12)
root.left.left = newnode(3)
# Populates nextRight pointer
# in all nodes
p = populateNext(root)
# Let us see the populated values
ptr = root.left.left
while(ptr != None):
out = 0
if(ptr.next != None):
out = ptr.next.data
else:
out = -1
# -1 is printed if there is no successor
print("Next of", ptr.data, "is", out)
ptr = ptr.next
# This code is contributed by Arnab Kundu
C#
// C# program to populate inorder traversal of all nodes
using System;
class BinaryTree
{
// A binary tree node
class Node
{
public int data;
public Node left, right, next;
public Node(int item)
{
data = item;
left = right = next = null;
}
}
Node root;
static Node next = null;
/* Set next of p and all descendants of p by traversing them in
reverse Inorder */
void populateNext(Node node)
{
// The first visited node will be the rightmost node
// next of the rightmost node will be NULL
if (node != null)
{
// First set the next pointer in right subtree
populateNext(node.right);
// Set the next as previously visited node in reverse Inorder
node.next = next;
// Change the prev for subsequent node
next = node;
// Finally, set the next pointer in left subtree
populateNext(node.left);
}
}
/* Driver program to test above functions*/
static public void Main(String []args )
{
/* Constructed binary tree is
10
/ \
8 12
/
3 */
BinaryTree tree = new BinaryTree();
tree.root = new Node(10);
tree.root.left = new Node(8);
tree.root.right = new Node(12);
tree.root.left.left = new Node(3);
// Populates nextRight pointer in all nodes
tree.populateNext(tree.root);
// Let us see the populated values
Node ptr = tree.root.left.left;
while (ptr != null)
{
// -1 is printed if there is no successor
int print = ptr.next != null ? ptr.next.data : -1;
Console.WriteLine("Next of " + ptr.data + " is: " + print);
ptr = ptr.next;
}
}
}
// This code has been contributed by Arnab Kundu
Javascript
<script>
// Javascript program to populate inorder traversal of all nodes
// A binary tree node
class Node
{
constructor(x) {
this.data = x;
this.left = null;
this.right = null;
}
}
let root;
let next = null;
/* Set next of p and all descendants of p by traversing them in
reverse Inorder */
function populateNext(node)
{
// The first visited node will be the rightmost node
// next of the rightmost node will be NULL
if (node != null)
{
// First set the next pointer in right subtree
populateNext(node.right);
// Set the next as previously visited node in reverse Inorder
node.next = next;
// Change the prev for subsequent node
next = node;
// Finally, set the next pointer in left subtree
populateNext(node.left);
}
}
/* Driver program to test above functions*/
/* Constructed binary tree is
10
/ \
8 12
/
3 */
root = new Node(10)
root.left = new Node(8)
root.right = new Node(12)
root.left.left = new Node(3)
// Populates nextRight pointer
// in all nodes
p = populateNext(root)
// Let us see the populated values
ptr = root.left.left
while (ptr != null)
{
// -1 is printed if there is no successor
let print = ptr.next != null ? ptr.next.data : -1;
document.write("Next of " + ptr.data + " is: " + print+"<br>");
ptr = ptr.next;
}
// This code is contributed by unknown2108
</script>
Output:
Next of 3 is 8 Next of 8 is 10 Next of 10 is 12 Next of 12 is -1
We can avoid the use of static variables by passing reference to next as a parameter.
C++
// An implementation that doesn't use static variable
// A wrapper over populateNextRecur
void populateNext(node *root)
{
// The first visited node will be the rightmost node
// next of the rightmost node will be NULL
node *next = NULL;
populateNextRecur(root, &next);
}
/* Set next of all descendants of p by
traversing them in reverse Inorder */
void populateNextRecur(node* p, node **next_ref)
{
if (p)
{
// First set the next pointer in right subtree
populateNextRecur(p->right, next_ref);
// Set the next as previously visited
// node in reverse Inorder
p->next = *next_ref;
// Change the prev for subsequent node
*next_ref = p;
// Finally, set the next pointer in right subtree
populateNextRecur(p->left, next_ref);
}
}
// This is code is contributed by rathbhupendra
C
// An implementation that doesn't use static variable
// A wrapper over populateNextRecur
void populateNext(struct node *root)
{
// The first visited node will be the rightmost node
// next of the rightmost node will be NULL
struct node *next = NULL;
populateNextRecur(root, &next);
}
/* Set next of all descendants of p by traversing them in reverse Inorder */
void populateNextRecur(struct node* p, struct node **next_ref)
{
if (p)
{
// First set the next pointer in right subtree
populateNextRecur(p->right, next_ref);
// Set the next as previously visited node in reverse Inorder
p->next = *next_ref;
// Change the prev for subsequent node
*next_ref = p;
// Finally, set the next pointer in right subtree
populateNextRecur(p->left, next_ref);
}
}
Java
// A wrapper over populateNextRecur
void populateNext(Node node) {
// The first visited node will be the rightmost node
// next of the rightmost node will be NULL
populateNextRecur(node, next);
}
/* Set next of all descendants of p by traversing them in reverse Inorder */
void populateNextRecur(Node p, Node next_ref) {
if (p != null) {
// First set the next pointer in right subtree
populateNextRecur(p.right, next_ref);
// Set the next as previously visited node in reverse Inorder
p.next = next_ref;
// Change the prev for subsequent node
next_ref = p;
// Finally, set the next pointer in right subtree
populateNextRecur(p.left, next_ref);
}
}
Python3
# A wrapper over populateNextRecur
def populateNext(node):
# The first visited node will be the rightmost node
# next of the rightmost node will be NULL
populateNextRecur(node, next)
# /* Set next of all descendants of p by
# traversing them in reverse Inorder */
def populateNextRecur(p, next_ref):
if (p != None):
# First set the next pointer in right subtree
populateNextRecur(p.right, next_ref)
# Set the next as previously visited node in reverse Inorder
p.next = next_ref
# Change the prev for subsequent node
next_ref = p
# Finally, set the next pointer in right subtree
populateNextRecur(p.left, next_ref)
# This code is contributed by Mohit kumar 29
C#
// A wrapper over populateNextRecur
void populateNext(Node node)
{
// The first visited node will be the rightmost node
// next of the rightmost node will be NULL
populateNextRecur(node, next);
}
/* Set next of all descendants of p by
traversing them in reverse Inorder */
void populateNextRecur(Node p, Node next_ref)
{
if (p != null)
{
// First set the next pointer in right subtree
populateNextRecur(p.right, next_ref);
// Set the next as previously visited node in reverse Inorder
p.next = next_ref;
// Change the prev for subsequent node
next_ref = p;
// Finally, set the next pointer in right subtree
populateNextRecur(p.left, next_ref);
}
}
// This code is contributed by princiraj1992
Javascript
<script>
// A wrapper over populateNextRecur
function populateNext(node)
{
// The first visited node will be the rightmost node
// next of the rightmost node will be NULL
populateNextRecur(node, next);
}
/* Set next of all descendants of p by
traversing them in reverse Inorder */
function populateNextRecur(p, next_ref)
{
if (p != null)
{
// First set the next pointer in right subtree
populateNextRecur(p.right, next_ref);
// Set the next as previously visited node in reverse Inorder
p.next = next_ref;
// Change the prev for subsequent node
next_ref = p;
// Finally, set the next pointer in right subtree
populateNextRecur(p.left, next_ref);
}
}
// This code is contributed by importantly.
</script>
Time Complexity: O(n)
Approach:
Steps:
- Create an array or an ArrayList.
- Store the inorder traversal of the binary tree into the ArrayList (nodes are to be stored).
- Now traverse the array and replace the next pointer of the node to the immediate right node(next element in the array which is the required inorder successor).
list.get(i).next = list.get(i+1)
Implementation:
Java
import java.util.ArrayList;
//class Node
class Node{
int data;
Node left,right,next;
//constructor for initializing key value and all the
//pointers
Node(int data){
this.data = data;
left = right = next = null;
}
}
public class Solution {
Node root = null;
//list to store inorder sequence
ArrayList<Node> list = new ArrayList<>();
//function for populating next pointer to inorder successor
void populateNext() {
//list = [3,8,10,12]
//inorder successor of the present node is the immediate
//right node
//foe ex: inorder successor of 3 is 8
for(int i=0;i<list.size();i++) {
//check if it is the last node
//point next of last node(right most) to null
if(i != list.size()-1) {
list.get(i).next = list.get(i+1);
}
else {
list.get(i).next = null;
}
}
// Let us see the populated values
Node ptr = root.left.left;
while (ptr != null)
{
// -1 is printed if there is no successor
int print = ptr.next != null ? ptr.next.data : -1;
System.out.println("Next of " + ptr.data + " is: " + print);
ptr = ptr.next;
}
}
//insert the inorder into a list to keep track
//of the inorder successor
void inorder(Node root) {
if(root!=null) {
inorder(root.left);
list.add(root);
inorder(root.right);
}
}
//Driver function
public static void main(String args[]) {
Solution tree = new Solution();
/* 10
/ \
8 12
/
3 */
tree.root = new Node(10);
tree.root.left = new Node(8);
tree.root.right = new Node(12);
tree.root.left.left = new Node(3);
//function calls
tree.inorder(tree.root);
tree.populateNext();
}
}
C#
using System;
using System.Collections.Generic;
// class Node
public
class Node {
public
int data;
public
Node left, right, next;
// constructor for initializing key value and all the
// pointers
public
Node(int data) {
this.data = data;
left = right = next = null;
}
}
public class Solution {
Node root = null;
// list to store inorder sequence
List<Node> list = new List<Node>();
// function for populating next pointer to inorder successor
void populateNext() {
// list = [3,8,10,12]
// inorder successor of the present node is the immediate
// right node
// foe ex: inorder successor of 3 is 8
for (int i = 0; i < list.Count; i++) {
// check if it is the last node
// point next of last node(right most) to null
if (i != list.Count - 1) {
list[i].next = list[i + 1];
} else {
list[i].next = null;
}
}
// Let us see the populated values
Node ptr = root.left.left;
while (ptr != null) {
// -1 is printed if there is no successor
int print = ptr.next != null ? ptr.next.data : -1;
Console.WriteLine("Next of " + ptr.data + " is: " + print);
ptr = ptr.next;
}
}
// insert the inorder into a list to keep track
// of the inorder successor
void inorder(Node root) {
if (root != null) {
inorder(root.left);
list.Add(root);
inorder(root.right);
}
}
// Driver function
public static void Main(String []args) {
Solution tree = new Solution();
/*
* 10 / \ 8 12 / 3
*/
tree.root = new Node(10);
tree.root.left = new Node(8);
tree.root.right = new Node(12);
tree.root.left.left = new Node(3);
// function calls
tree.inorder(tree.root);
tree.populateNext();
}
}
// This code is contributed by gauravrajput1
Javascript
<script>
// class Node
class Node
{
// constructor for initializing key value and all the
// pointers
constructor(data)
{
this.data = data;
this.left = this.right = this.next = null;
}
}
let root = null;
// list to store inorder sequence
let list = [];
// function for populating next pointer to inorder successor
function populateNext()
{
// list = [3,8,10,12]
// inorder successor of the present node is the immediate
// right node
// foe ex: inorder successor of 3 is 8
for(let i = 0; i < list.length; i++)
{
// check if it is the last node
// point next of last node(right most) to null
if(i != list.length - 1)
{
list[i].next = list[i + 1];
}
else {
list[i].next = null;
}
}
// Let us see the populated values
let ptr = root.left.left;
while (ptr != null)
{
// -1 is printed if there is no successor
let print = ptr.next != null ? ptr.next.data : -1;
document.write("Next of " + ptr.data + " is: " + print+"<br>");
ptr = ptr.next;
}
}
// insert the inorder into a list to keep track
// of the inorder successor
function inorder(root)
{
if(root != null)
{
inorder(root.left);
list.push(root);
inorder(root.right);
}
}
// Driver function
/* 10
/ \
8 12
/
3 */
root = new Node(10);
root.left = new Node(8);
root.right = new Node(12);
root.left.left = new Node(3);
// function calls
inorder(root);
populateNext();
// This code is contributed by avanitrachhadiya2155
</script>
Output
Next of 3 is: 8 Next of 8 is: 10 Next of 10 is: 12 Next of 12 is: -1
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