Level Order Successor of a node in Binary Tree

Given a binary tree and a node in the binary tree, find Levelorder successor of the given node. That is, the node that appears after the given node in the level order traversal of the tree.

Note: The task is not just to print the data of the node, you have to return the complete node from the tree.

Examples: 

Consider the following binary tree
              20            
           /      \         
          10       26       
         /  \     /   \     
       4     18  24    27   
            /  \
           14   19
          /  \
         13  15

Levelorder traversal of given tree is:
20, 10, 26, 4, 18, 24, 27, 14, 19, 13, 15

Input : 24
Output : 27

Input : 4
Output : 18 

Approach:  

1. Check if the root is NULL, that is tree is empty. If true then return NULL.
2. Check if the given node is root. If true: 
   • Check if left child of root exists, if true return left child of root.
   • Else, check if right child exists, return it.
   • If the root is the only node. Return NULL.
3. Otherwise, perform Level Order Traversal on the tree using a Queue.
4. At every step of the level order traversal, check if the current node matches with the given node.
5. If True, stop traversing any further and return the element at top of queue which will be the next node in the level order traversal.

Below is the implementation of the above approach: 

C++

// CPP program to find Levelorder
// successor of given node in the
// Binary Tree

#include <bits/stdc++.h>
using namespace std;

// Tree Node
struct Node {
    struct Node *left, *right;
    int value;
};

// Utility function to create a
// new node with given value
struct Node* newNode(int value)
{
    Node* temp = new Node;
    temp->left = temp->right = NULL;
    temp->value = value;

    return temp;
}

// Function to find the Level Order Successor
// of a given Node in Binary Tree
Node* levelOrderSuccessor(Node* root, Node* key)
{
    // Base Case
    if (root == NULL)
        return NULL;

    // If root equals to key
    if (root == key) {

        // If left child exists it will be
        // the Postorder Successor
        if (root->left)
            return root->left;

        // Else if right child exists it will be
        // the Postorder Successor
        else if (root->right)
            return root->right;
        else
            return NULL; // No Successor
    }

    // Create an empty queue for level
    // order traversal
    queue<Node*> q;

    // Enqueue Root
    q.push(root);

    while (!q.empty()) {
        Node* nd = q.front();
        q.pop();

        if (nd->left != NULL) {
            q.push(nd->left);
        }

        if (nd->right != NULL) {
            q.push(nd->right);
        }

        if (nd == key)
            break;
    }

    return q.front();
}

// Driver code
int main()
{
    struct Node* root = newNode(20);
    root->left = newNode(10);
    root->left->left = newNode(4);
    root->left->right = newNode(18);
    root->right = newNode(26);
    root->right->left = newNode(24);
    root->right->right = newNode(27);
    root->left->right->left = newNode(14);
    root->left->right->left->left = newNode(13);
    root->left->right->left->right = newNode(15);
    root->left->right->right = newNode(19);

    struct Node* key = root->right->left; // node 24

    struct Node* res = levelOrderSuccessor(root, key);

    if (res)
        cout << "LevelOrder successor of "
             << key->value << " is " << res->value;
    else
        cout << "LevelOrder successor of "
             << key->value << " is "  << "NULL";

    return 0;
}

Java

// Java program to find Levelorder
// successor of given node in the
// Binary Tree
import java.util.*;
class GfG {

// Tree Node
static class Node {
    Node left, right;
    int value;
}

// Utility function to create a
// new node with given value
static Node newNode(int value)
{
    Node temp = new Node();
    temp.left = null;
    temp.right = null;
    temp.value = value;

    return temp;
}

// Function to find the Level Order Successor
// of a given Node in Binary Tree
static Node levelOrderSuccessor(Node root, Node key)
{
    // Base Case
    if (root == null)
        return null;

    // If root equals to key
    if (root == key) {

        // If left child exists it will be
        // the Postorder Successor
        if (root.left != null)
            return root.left;

        // Else if right child exists it will be
        // the Postorder Successor
        else if (root.right != null)
            return root.right;
        else
            return null; // No Successor
    }

    // Create an empty queue for level
    // order traversal
    Queue<Node> q = new LinkedList<Node> ();

    // Enqueue Root
    q.add(root);

    while (!q.isEmpty()) {
        Node nd = q.peek();
        q.remove();

        if (nd.left != null) {
            q.add(nd.left);
        }

        if (nd.right != null) {
            q.add(nd.right);
        }

        if (nd == key)
            break;
    }

    return q.peek();
}

// Driver code
public static void main(String[] args)
{
    Node root = newNode(20);
    root.left = newNode(10);
    root.left.left = newNode(4);
    root.left.right = newNode(18);
    root.right = newNode(26);
    root.right.left = newNode(24);
    root.right.right = newNode(27);
    root.left.right.left = newNode(14);
    root.left.right.left.left = newNode(13);
    root.left.right.left.right = newNode(15);
    root.left.right.right = newNode(19);

    Node key = root.right.left; // node 24

Node res = levelOrderSuccessor(root, key);

    if (res != null)
        System.out.println("LevelOrder successor of "
                        +key.value + " is " + res.value);
    else
        System.out.println("LevelOrder successor of "
                            +key.value + " is NULL");

}
}

Python3

# Python3 program to find Level
# order successor of given node
# in the Binary Tree

# Node definition
class Node:

    def __init__(self, value):
        self.left = None
        self.right = None
        self.value = value

# Function to find the Level
# Order Successor of a given
# Node in Binary Tree
def levelOrderSuccessor(root, key):

    # Base Case
    if root == None:
        return None

    # If root equals to key
    elif root == key:

        # If left child exists, it will
        # be the PostOrder Successor
        if root.left:
            return root.left

        # Else if right child exists, it
        # will be the PostOrder Successor
        elif root.right:
            return root.right

        # No Successor
        else:
            return None

    # Create an empty queue for
    # level order traversal
    q = []

    # Enqueue Root
    q.append(root)

    while len(q) != 0:
        nd = q.pop(0)

        if nd.left != None:
            q.append(nd.left)

        if nd.right != None:
            q.append(nd.right)

        if nd == key:
            break

    return q[0]

# Driver Code
if __name__ == "__main__":

    root = Node(20)
    root.left = Node(10)
    root.left.left = Node(4)
    root.left.right = Node(18)
    root.right = Node(26)
    root.right.left = Node(24)
    root.right.right = Node(27)
    root.left.right.left = Node(14)
    root.left.right.left.left = Node(13)
    root.left.right.left.right = Node(15)
    root.left.right.right = Node(19)

    key = root.right.left # node 24

    res = levelOrderSuccessor(root, key)

    if res:
        print("LevelOrder successor of " +
                 str(key.value) + " is " +
                 str(res.value))

    else:
        print("LevelOrder successor of " +
              str(key.value) + " is NULL")

# This code is contributed
# by Rituraj Jain

C#

// C# program to find Levelorder
// successor of given node in the
// Binary Tree
using System;
using System.Collections.Generic;

class GfG
{

// Tree Node
public class Node
{
    public Node left, right;
    public int value;
}

// Utility function to create a
// new node with given value
static Node newNode(int value)
{
    Node temp = new Node();
    temp.left = null;
    temp.right = null;
    temp.value = value;

    return temp;
}

// Function to find the Level Order Successor
// of a given Node in Binary Tree
static Node levelOrderSuccessor(Node root, Node key)
{
    // Base Case
    if (root == null)
        return null;

    // If root equals to key
    if (root == key)
    {

        // If left child exists it will be
        // the Postorder Successor
        if (root.left != null)
            return root.left;

        // Else if right child exists it will be
        // the Postorder Successor
        else if (root.right != null)
            return root.right;
        else
            return null; // No Successor
    }

    // Create an empty queue for level
    // order traversal
    LinkedList<Node> q = new LinkedList<Node> ();

    // Enqueue Root
    q.AddLast(root);

    while (q.Count != 0)
    {
        Node nd = q.First.Value;
        q.RemoveFirst();

        if (nd.left != null)
        {
            q.AddLast(nd.left);
        }

        if (nd.right != null)
        {
            q.AddLast(nd.right);
        }

        if (nd == key)
            break;
    }

    return q.First.Value;
}

// Driver code
public static void Main(String[] args)
{
    Node root = newNode(20);
    root.left = newNode(10);
    root.left.left = newNode(4);
    root.left.right = newNode(18);
    root.right = newNode(26);
    root.right.left = newNode(24);
    root.right.right = newNode(27);
    root.left.right.left = newNode(14);
    root.left.right.left.left = newNode(13);
    root.left.right.left.right = newNode(15);
    root.left.right.right = newNode(19);

    Node key = root.right.left; // node 24

    Node res = levelOrderSuccessor(root, key);

    if (res != null)
        Console.WriteLine("LevelOrder successor of "
                        +key.value + " is " + res.value);
    else
        Console.WriteLine("LevelOrder successor of "
                            +key.value + " is NULL");

}
}

// This code has been contributed by 29AjayKumar

Javascript

<script>

    // JavaScript program to find Levelorder
    // successor of given node in the
    // Binary Tree

    // Tree Node
    class Node
    {
        constructor(value) {
           this.left = null;
           this.right = null;
           this.value = value;
        }
    }

    // Utility function to create a
    // new node with given value
    function newNode(value)
    {
        let temp = new Node(value);
        return temp;
    }

    // Function to find the Level Order Successor
    // of a given Node in Binary Tree
    function levelOrderSuccessor(root, key)
    {
        // Base Case
        if (root == null)
            return null;

        // If root equals to key
        if (root == key) {

            // If left child exists it will be
            // the Postorder Successor
            if (root.left != null)
                return root.left;

            // Else if right child exists it will be
            // the Postorder Successor
            else if (root.right != null)
                return root.right;
            else
                return null; // No Successor
        }

        // Create an empty queue for level
        // order traversal
        let q = [];

        // Enqueue Root
        q.push(root);

        while (q.length > 0) {
            let nd = q[0];
            q.shift();

            if (nd.left != null) {
                q.push(nd.left);
            }

            if (nd.right != null) {
                q.push(nd.right);
            }

            if (nd == key)
                break;
        }

        return q[0];
    }

    let root = newNode(20);
    root.left = newNode(10);
    root.left.left = newNode(4);
    root.left.right = newNode(18);
    root.right = newNode(26);
    root.right.left = newNode(24);
    root.right.right = newNode(27);
    root.left.right.left = newNode(14);
    root.left.right.left.left = newNode(13);
    root.left.right.left.right = newNode(15);
    root.left.right.right = newNode(19);

    let key = root.right.left; // node 24

    let res = levelOrderSuccessor(root, key);

    if (res != null)
      document.write("LevelOrder successor of "
                         +key.value + " is " + res.value);
    else
      document.write("LevelOrder successor of "
                         +key.value + " is NULL");

</script>

Output: 

LevelOrder successor of 24 is 27

Time Complexity: O(N), as we are using a while loop which will traverse N times, where N is the number of nodes in the tree.

Auxiliary Space: O(N), as we are using extra space for the queue, which we are using for the level order traversal.