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 5Solution: 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 Nodestruct 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 levelnode* 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 nodenode* newNode(int key){ node *temp = new node; temp->key = key; temp->left = temp->right = NULL; return temp;} // A utility function to test above functionsvoid 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 functionsint 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 nodeclass 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 Nodeclass 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 leveldef 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 functionroot = 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.
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