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
Input : 2
Output : 6
Input : 4
Output : 5
In our previous post we have discussed about a solution using Level Order Traversal. In this post we will discuss about a solution based on Preorder traversal which takes constant auxiliary space.
The idea is to traverse the given tree using preorder traversal and search for the given key. Once we found the given key, we will mark the level number for this key. Now the next node we will find at the same level is the required node which is at the right of given key.
Below is the implementation of above idea:
C++
/* C++ program to find next right of a given key using preorder traversal */#include <bits/stdc++.h>using namespace std;// A Binary Tree Nodestruct Node { struct Node *left, *right; int key;};// 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;}// Function to find next node for given node// in same level in a binary tree by using// pre-order traversalNode* nextRightNode(Node* root, int k, int level, int& value_level){ // return null if tree is empty if (root == NULL) return NULL; // if desired node is found, set value_level // to current level if (root->key == k) { value_level = level; return NULL; } // if value_level is already set, then current // node is the next right node else if (value_level) { if (level == value_level) return root; } // recurse for left subtree by increasing level by 1 Node* leftNode = nextRightNode(root->left, k, level + 1, value_level); // if node is found in left subtree, return it if (leftNode) return leftNode; // recurse for right subtree by increasing level by 1 return nextRightNode(root->right, k, level + 1, value_level);}// Function to find next node of given node in the// same level in given binary treeNode* nextRightNodeUtil(Node* root, int k){ int value_level = 0; return nextRightNode(root, k, 1, value_level);}// A utility function to test above functionsvoid test(Node* root, int k){ Node* nr = nextRightNodeUtil(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 using preorder traversal */import java.util.*;class GfG { // A Binary Tree Node static class Node { Node left, right; int key; }// Utility function to create a new tree node static Node newNode(int key) { Node temp = new Node(); temp.key = key; temp.left = null; temp.right = null; return temp; } // Function to find next node for given node // in same level in a binary tree by using // pre-order traversal static Node nextRightNode(Node root, int k, int level, int[] value) { // return null if tree is empty if (root == null) return null; // if desired node is found, set value[0] // to current level if (root.key == k) { value[0] = level; return null; } // if value[0] is already set, then current // node is the next right node else if (value[0] != 0) { if (level == value[0]) return root; } // recurse for left subtree by increasing level by 1 Node leftNode = nextRightNode(root.left, k, level + 1, value); // if node is found in left subtree, return it if (leftNode != null) return leftNode; // recurse for right subtree by increasing level by 1 return nextRightNode(root.right, k, level + 1, value); } // Function to find next node of given node in the // same level in given binary tree static Node nextRightNodeUtil(Node root, int k) { int[] v = new int[1]; v[0] = 0; return nextRightNode(root, k, 1, v); } // A utility function to test above functions static void test(Node root, int k) { Node nr = nextRightNodeUtil(root, k); if (nr != null) System.out.println("Next Right of " + k + " is "+ nr.key); else System.out.println("No next right node found for " + k);} // Driver program to test above functions public static void main(String[] args) { // 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); } } // This code has been contributed by Mukul Sharma |
Python3
# Python3 program to find next right of a # given key using preorder traversal # class to create a new tree node class newNode: def __init__(self, key): self.key = key self.left = self.right = None# Function to find next node for given node # in same level in a binary tree by using # pre-order traversal def nextRightNode(root, k, level, value_level): # return None if tree is empty if (root == None): return None # if desired node is found, set # value_level to current level if (root.key == k): value_level[0] = level return None # if value_level is already set, then # current node is the next right node elif (value_level[0]): if (level == value_level[0]): return root # recurse for left subtree by increasing # level by 1 leftNode = nextRightNode(root.left, k, level + 1, value_level) # if node is found in left subtree, # return it if (leftNode): return leftNode # recurse for right subtree by # increasing level by 1 return nextRightNode(root.right, k, level + 1, value_level)# Function to find next node of given node # in the same level in given binary tree def nextRightNodeUtil(root, k): value_level = [0] return nextRightNode(root, k, 1, value_level) # A utility function to test above functions def test(root, k): nr = nextRightNodeUtil(root, k) if (nr != None): print("Next Right of", k, "is", nr.key) else: print("No next right node found for", k)# Driver Codeif __name__ == '__main__': # Let us create binary tree given in the # above example 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)# This code is contributed by PranchalK |
C#
/* C# program to find next right of a given key using preorder traversal */using System;class GfG { public class V { public int value_level = 0; } // A Binary Tree Node public class Node{ public Node left, right; public int key; } // Utility function to create a new tree node static Node newNode(int key) { Node temp = new Node(); temp.key = key; temp.left = null; temp.right = null; return temp; } // Function to find next node for given node // in same level in a binary tree by using // pre-order traversal static Node nextRightNode(Node root, int k, int level, V value) { // return null if tree is empty if (root == null) return null; // if desired node is found, set // value_level to current level if (root.key == k) { value.value_level = level; return null; } // if value_level is already set, then current // node is the next right node else if (value.value_level != 0) { if (level == value.value_level) return root; } // recurse for left subtree by increasing level by 1 Node leftNode = nextRightNode(root.left, k, level + 1, value); // if node is found in left subtree, return it if (leftNode != null) return leftNode; // recurse for right subtree by // increasing level by 1 return nextRightNode(root.right, k, level + 1, value); } // Function to find next node of given node in the // same level in given binary tree static Node nextRightNodeUtil(Node root, int k) { V v = new V(); return nextRightNode(root, k, 1, v); } // A utility function to test above functions static void test(Node root, int k) { Node nr = nextRightNodeUtil(root, k); if (nr != null) Console.WriteLine("Next Right of " + k + " is "+ nr.key); else Console.WriteLine("No next right node" + " found for " + k); } // Driver mainpublic static void Main(String[] args) { // 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); } } // This code contributed by Rajput-Ji |
Javascript
<script> /* Javascript program to find next right of a given key using preorder traversal */ // A Binary Tree Node class Node { constructor(key) { this.left = null; this.right = null; this.key = key; } } // Utility function to create a new tree node function newNode(key) { let temp = new Node(key); return temp; } // Function to find next node for given node // in same level in a binary tree by using // pre-order traversal function nextRightNode(root, k, level, value) { // return null if tree is empty if (root == null) return null; // if desired node is found, set value[0] // to current level if (root.key == k) { value[0] = level; return null; } // if value[0] is already set, then current // node is the next right node else if (value[0] != 0) { if (level == value[0]) return root; } // recurse for left subtree by increasing level by 1 let leftNode = nextRightNode(root.left, k, level + 1, value); // if node is found in left subtree, return it if (leftNode != null) return leftNode; // recurse for right subtree by increasing level by 1 return nextRightNode(root.right, k, level + 1, value); } // Function to find next node of given node in the // same level in given binary tree function nextRightNodeUtil(root, k) { let v = new Array(1); v[0] = 0; return nextRightNode(root, k, 1, v); } // A utility function to test above functions function test(root, k) { let nr = nextRightNodeUtil(root, k); if (nr != null) document.write("Next Right of " + k + " is "+ nr.key + "</br>"); else document.write("No next right node found for " + k + "</br>"); } // Let us create binary tree given in the // above example let 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); </script> |
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: O(n)
Auxiliary Space: O(1)
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