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Populate Inorder Successor for all nodes

  • Difficulty Level : Medium
  • Last Updated : 01 Jul, 2021

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




struct node
{
  int data;
  struct node* left;
  struct node* right;
  struct node* next;
}

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 descendents 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:

  1. Create an array or an ArrayList.
  2. Store the inorder traversal of the binary tree into the ArrayList (nodes are to be stored).
  3. 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 succesoor 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();
    }
}

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 succesoor 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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