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 in-order 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

Time Complexity: O(n)
Auxiliary Space : O(1)