Find next right node of a given key

Given a Binary tree and a key in the binary tree, find the node right to the given key. If there is no node on right side, then return NULL. Expected time complexity is O(n) where n is the number of nodes in the given binary tree.

For example, consider the following Binary Tree. Output for 2 is 6, output for 4 is 5. Output for 10, 6 and 5 is NULL.

                  10
               /      \
          2         6
           /   \         \ 
     8      4          5

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Solution: The idea is to do level order traversal of given Binary Tree. When we find the given key, we just check if the next node in level order traversal is of same level, if yes, we return the next node, otherwise return NULL.

C++

/* Program to find next right of a given key */
#include <iostream> 
#include <queue> 
using namespace std; 
  
// A Binary Tree Node 
struct node 
{ 
    struct node *left, *right; 
    int key; 
}; 
  
// Method to find next right of given key k, it returns NULL if k is 
// not present in tree or k is the rightmost node of its level 
node* nextRight(node *root, int k) 
{ 
    // Base Case 
    if (root == NULL) 
        return 0; 
  
    // Create an empty queue for level order tarversal 
    queue<node *> qn; // A queue to store node addresses 
    queue<int> ql;   // Another queue to store node levels 
  
    int level = 0;  // Initialize level as 0 
  
    // Enqueue Root and its level 
    qn.push(root); 
    ql.push(level); 
  
    // A standard BFS loop 
    while (qn.size()) 
    { 
        // dequeue an node from qn and its level from ql 
        node *node = qn.front(); 
        level = ql.front(); 
        qn.pop(); 
        ql.pop(); 
  
        // If the dequeued node has the given key k 
        if (node->key == k) 
        { 
            // If there are no more items in queue or given node is 
            // the rightmost node of its level, then return NULL 
            if (ql.size() == 0 || ql.front() != level) 
               return NULL; 
  
            // Otherwise return next node from queue of nodes 
            return qn.front(); 
        } 
  
        // Standard BFS steps: enqueue children of this node 
        if (node->left != NULL) 
        { 
            qn.push(node->left); 
            ql.push(level+1); 
        } 
        if (node->right != NULL) 
        { 
            qn.push(node->right); 
            ql.push(level+1); 
        } 
    } 
  
    // We reach here if given key x doesn't exist in tree 
    return NULL; 
} 
  
// Utility function to create a new tree node 
node* newNode(int key) 
{ 
    node *temp = new node; 
    temp->key = key; 
    temp->left = temp->right = NULL; 
    return temp; 
} 
  
// A utility function to test above functions 
void test(node *root, int k) 
{ 
    node *nr = nextRight(root, k); 
    if (nr != NULL) 
      cout << "Next Right of " << k << " is " << nr->key << endl; 
    else
      cout << "No next right node found for " << k << endl; 
} 
  
// Driver program to test above functions 
int main() 
{ 
    // Let us create binary tree given in the above example 
    node *root = newNode(10); 
    root->left = newNode(2); 
    root->right = newNode(6); 
    root->right->right = newNode(5); 
    root->left->left = newNode(8); 
    root->left->right = newNode(4); 
  
    test(root, 10); 
    test(root, 2); 
    test(root, 6); 
    test(root, 5); 
    test(root, 8); 
    test(root, 4); 
    return 0; 
} 

Java

// Java program to find next right of a given key 
   
import java.util.LinkedList; 
import java.util.Queue; 
   
// A binary tree node 
class Node  
{ 
    int data; 
    Node left, right; 
   
    Node(int item)  
    { 
        data = item; 
        left = right = null; 
    } 
} 
   
class BinaryTree  
{ 
    Node root; 
   
    // Method to find next right of given key k, it returns NULL if k is 
    // not present in tree or k is the rightmost node of its level 
    Node nextRight(Node first, int k)  
    { 
        // Base Case 
        if (first == null) 
            return null; 
   
        // Create an empty queue for level order tarversal 
        // A queue to store node addresses 
        Queue<Node> qn = new LinkedList<Node>();  
          
        // Another queue to store node levels 
        Queue<Integer> ql = new LinkedList<Integer>();    
   
        int level = 0;  // Initialize level as 0 
   
        // Enqueue Root and its level 
        qn.add(first); 
        ql.add(level); 
   
        // A standard BFS loop 
        while (qn.size() != 0)  
        { 
            // dequeue an node from qn and its level from ql 
            Node node = qn.peek(); 
            level = ql.peek(); 
            qn.remove(); 
            ql.remove(); 
   
            // If the dequeued node has the given key k 
            if (node.data == k)  
            { 
                // If there are no more items in queue or given node is 
                // the rightmost node of its level, then return NULL 
                if (ql.size() == 0 || ql.peek() != level) 
                    return null; 
   
                // Otherwise return next node from queue of nodes 
                return qn.peek(); 
            } 
   
            // Standard BFS steps: enqueue children of this node 
            if (node.left != null)  
            { 
                qn.add(node.left); 
                ql.add(level + 1); 
            } 
            if (node.right != null)  
            { 
                qn.add(node.right); 
                ql.add(level + 1); 
            } 
        } 
   
        // We reach here if given key x doesn't exist in tree 
        return null; 
    } 
   
    // A utility function to test above functions 
    void test(Node node, int k)  
    { 
        Node nr = nextRight(root, k); 
        if (nr != null) 
            System.out.println("Next Right of " + k + " is " + nr.data); 
        else
            System.out.println("No next right node found for " + k); 
    } 
   
    // Driver program to test above functions 
    public static void main(String args[])  
    { 
        BinaryTree tree = new BinaryTree(); 
        tree.root = new Node(10); 
        tree.root.left = new Node(2); 
        tree.root.right = new Node(6); 
        tree.root.right.right = new Node(5); 
        tree.root.left.left = new Node(8); 
        tree.root.left.right = new Node(4); 
   
        tree.test(tree.root, 10); 
        tree.test(tree.root, 2); 
        tree.test(tree.root, 6); 
        tree.test(tree.root, 5); 
        tree.test(tree.root, 8); 
        tree.test(tree.root, 4); 
   
    } 
} 
   
// This code has been contributed by Mayank Jaiswal 

Python

# Python program to find next right node of given key 
  
# A Binary Tree Node 
class Node: 
      
    # Constructor to create a new node 
    def __init__(self, key): 
        self.key = key 
        self.left = None
        self.right = None
      
# Method to find next right of a given key k, it returns 
# None if k is not present in tree or k is the rightmost 
# node of its level 
def nextRight(root, k): 
  
    # Base Case 
    if root is None: 
        return 0 
  
    # Create an empty queue for level order traversal 
    qn =  [] # A queue to store node addresses 
    q1 = [] # Another queue to store node levels 
  
    level = 0
  
    # Enqueue root and its level 
    qn.append(root) 
    q1.append(level) 
  
    # Standard BFS loop 
    while(len(qn) > 0): 
  
        # Dequeu an node from qn and its level from q1 
        node = qn.pop(0) 
        level = q1.pop(0) 
  
        # If the dequeued node has the given key k 
        if node.key == k : 
  
            # If there are no more items in queue or given 
            # node is the rightmost node of its level,  
            # then return None 
            if (len(q1) == 0 or q1[0] != level): 
                return None
  
            # Otherwise return next node from queue of nodes 
            return qn[0] 
  
        # Standard BFS steps: enqueue children of this node 
        if node.left is not None: 
            qn.append(node.left) 
            q1.append(level+1) 
  
        if node.right is not None: 
            qn.append(node.right) 
            q1.append(level+1) 
  
    # We reach here if given key x doesn't exist in tree 
    return None
  
def test(root, k): 
    nr = nextRight(root, k) 
    if nr is not None: 
        print "Next Right of " + str(k) + " is " + str(nr.key) 
    else: 
        print "No next right node found for " + str(k) 
  
# Driver program to test above function 
root = Node(10) 
root.left = Node(2) 
root.right = Node(6) 
root.right.right = Node(5) 
root.left.left = Node(8) 
root.left.right = Node(4) 
  
test(root, 10) 
test(root, 2) 
test(root, 6) 
test(root, 5) 
test(root, 8) 
test(root, 4) 
  
# This code is contributed by Nikhil Kumar Singh(nickzuck_007) 

C#

// C# program to find next right of a given key  
using System; 
using System.Collections.Generic; 
  
// A binary tree node  
public class Node  
{  
    public int data;  
    public Node left, right;  
  
    public Node(int item)  
    {  
        data = item;  
        left = right = null;  
    }  
}  
  
public class BinaryTree  
{  
    Node root;  
  
    // Method to find next right of given 
    // key k, it returns NULL if k is  
    // not present in tree or k is the  
    // rightmost node of its level  
    Node nextRight(Node first, int k)  
    {  
        // Base Case  
        if (first == null)  
            return null;  
  
        // Create an empty queue for 
        // level order tarversal  
        // A queue to store node addresses  
        Queue<Node> qn = new Queue<Node>();  
          
        // Another queue to store node levels  
        Queue<int> ql = new Queue<int>();  
  
        int level = 0; // Initialize level as 0  
  
        // Enqueue Root and its level  
        qn.Enqueue(first);  
        ql.Enqueue(level);  
  
        // A standard BFS loop  
        while (qn.Count != 0)  
        {  
            // dequeue an node from  
            // qn and its level from ql  
            Node node = qn.Peek();  
            level = ql.Peek();  
            qn.Dequeue();  
            ql.Dequeue();  
  
            // If the dequeued node has the given key k  
            if (node.data == k)  
            {  
                // If there are no more items 
                // in queue or given node is  
                // the rightmost node of its  
                // level, then return NULL  
                if (ql.Count == 0 || ql.Peek() != level)  
                    return null;  
  
                // Otherwise return next  
                // node from queue of nodes  
                return qn.Peek();  
            }  
  
            // Standard BFS steps: enqueue 
            // children of this node  
            if (node.left != null)  
            {  
                qn.Enqueue(node.left);  
                ql.Enqueue(level + 1);  
            }  
            if (node.right != null)  
            {  
                qn.Enqueue(node.right);  
                ql.Enqueue(level + 1);  
            }  
        }  
  
        // We reach here if given  
        // key x doesn't exist in tree  
        return null;  
    }  
  
    // A utility function to test above functions  
    void test(Node node, int k)  
    {  
        Node nr = nextRight(root, k);  
        if (nr != null)  
            Console.WriteLine("Next Right of " +  
                        k + " is " + nr.data);  
        else
            Console.WriteLine("No next right node found for " + k);  
    }  
  
    // Driver code  
    public static void Main(String []args)  
    {  
        BinaryTree tree = new BinaryTree();  
        tree.root = new Node(10);  
        tree.root.left = new Node(2);  
        tree.root.right = new Node(6);  
        tree.root.right.right = new Node(5);  
        tree.root.left.left = new Node(8);  
        tree.root.left.right = new Node(4);  
  
        tree.test(tree.root, 10);  
        tree.test(tree.root, 2);  
        tree.test(tree.root, 6);  
        tree.test(tree.root, 5);  
        tree.test(tree.root, 8);  
        tree.test(tree.root, 4);  
    }  
}  
  
// This code has been contributed 
// by 29AjayKumar 

Output:

No next right node found for 10
Next Right of 2 is 6
No next right node found for 6
No next right node found for 5
Next Right of 8 is 4
Next Right of 4 is 5

Time Complexity: The above code is a simple BFS traversal code which visits every enqueue and dequeues a node at most once. Therefore, the time complexity is O(n) where n is the number of nodes in the given binary tree.

Exercise: Write a function to find left node of a given node. If there is no node on the left side, then return NULL.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python

C#

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