Given a binary tree and an integer k, the task is to print the kth node in the vertical order traversal of binary tree.If no such node exists then print -1.
The vertical order traversal of a binary tree means to print it vertically.
Examples:
Input: 1 / \ 2 3 / \ / \ 4 5 6 7 \ \ 8 9 k = 3 Output: 1 The vertical order traversal of above tree is: 4 2 1 5 6 3 8 7 9 Input: 1 / \ 2 3 / \ / \ 4 5 6 7 \ \ 8 9 k = 13 Output: -1
Approach: The idea is to perform vertical order traversal and check if the current node is the kth node then print its value, if number of nodes in the tree is less than K then print -1.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Structure for a binary tree node struct Node { int key; Node *left, *right; }; // A utility function to create a new node Node* newNode( int key) { Node* node = new Node; node->key = key; node->left = node->right = NULL; return node; } // Function to find kth node // in vertical order traversal int KNodeVerticalOrder(Node* root, int k) { // Base case if (!root || k == 0) return -1; int n = 0; // Variable to store kth node int k_node = -1; // Create a map and store vertical order in // map map< int , vector< int > > m; int hd = 0; // Create queue to do level order traversal // Every item of queue contains node and // horizontal distance queue<pair<Node*, int > > que; que.push(make_pair(root, hd)); while (!que.empty()) { // Pop from queue front pair<Node*, int > temp = que.front(); que.pop(); hd = temp.second; Node* node = temp.first; // Insert this node's data in vector of hash m[hd].push_back(node->key); if (node->left != NULL) que.push(make_pair(node->left, hd - 1)); if (node->right != NULL) que.push(make_pair(node->right, hd + 1)); } // Traverse the map and find kth // node map< int , vector< int > >::iterator it; for (it = m.begin(); it != m.end(); it++) { for ( int i = 0; i < it->second.size(); ++i) { n++; if (n == k) return (it->second[i]); } } if (k_node == -1) return -1; } // Driver code int main() { Node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); root->right->left = newNode(6); root->right->right = newNode(7); root->right->left->right = newNode(8); root->right->right->right = newNode(9); root->right->right->left = newNode(10); root->right->right->left->right = newNode(11); root->right->right->left->right->right = newNode(12); int k = 5; cout << KNodeVerticalOrder(root, k); return 0; } |
Java
// Java implementation of the approach import java.util.ArrayList; import java.util.HashMap; import java.util.LinkedList; import java.util.Map; import java.util.Queue; import java.util.SortedMap; import java.util.TreeMap; class GFG{ // Structure for a binary tree node static class Node { int key; Node left, right; public Node( int key) { this .key = key; this .left = this .right = null ; } }; static class Pair { Node first; int second; public Pair(Node first, int second) { this .first = first; this .second = second; } } // Function to find kth node // in vertical order traversal static int KNodeVerticalOrder(Node root, int k) { // Base case if (root == null || k == 0 ) return - 1 ; int n = 0 ; // Variable to store kth node int k_node = - 1 ; // Create a map and store vertical order in // map SortedMap<Integer, ArrayList<Integer>> m = new TreeMap<Integer, ArrayList<Integer>>(); int hd = 0 ; // Create queue to do level order traversal // Every item of queue contains node and // horizontal distance Queue<Pair> que = new LinkedList<>(); que.add( new Pair(root, hd)); while (!que.isEmpty()) { // Pop from queue front Pair temp = que.poll(); hd = temp.second; Node node = temp.first; // Insert this node's data in // vector of hash if (m.get(hd) == null ) m.put(hd, new ArrayList<>()); m.get(hd).add(node.key); if (node.left != null ) que.add( new Pair(node.left, hd - 1 )); if (node.right != null ) que.add( new Pair(node.right, hd + 1 )); } // Traverse the map and find kth // node for (Map.Entry<Integer, ArrayList<Integer>> it : m.entrySet()) { for ( int i = 0 ; i < it.getValue().size(); i++) { n++; if (n == k) { return it.getValue().get(i); } } } if (k_node == - 1 ) return - 1 ; return 0 ; } // Driver code public static void main(String[] args) { Node root = new Node( 1 ); root.left = new Node( 2 ); root.right = new Node( 3 ); root.left.left = new Node( 4 ); root.left.right = new Node( 5 ); root.right.left = new Node( 6 ); root.right.right = new Node( 7 ); root.right.left.right = new Node( 8 ); root.right.right.right = new Node( 9 ); root.right.right.left = new Node( 10 ); root.right.right.left.right = new Node( 11 ); root.right.right.left.right.right = new Node( 12 ); int k = 5 ; System.out.println(KNodeVerticalOrder(root, k)); } } // This code is contributed by sanjeev2552 |
Python3
# Python3 implementation of the approach # Tree node structure class Node: def __init__( self , key): self .key = key self .left = None self .right = None # Function to find kth node # in vertical order traversal def KNodeVerticalOrder(root, k): # Base case if not root or k = = 0 : return - 1 n = 0 # Variable to store kth node k_node = - 1 # Create a map and store # vertical order in map m = {} hd = 0 # Create queue to do level order # traversal Every item of queue contains # node and horizontal distance que = [] que.append((root, hd)) while len (que) > 0 : # Pop from queue front temp = que.pop( 0 ) hd = temp[ 1 ] node = temp[ 0 ] # Insert this node's data in vector of hash if hd not in m: m[hd] = [] m[hd].append(node.key) if node.left ! = None : que.append((node.left, hd - 1 )) if node.right ! = None : que.append((node.right, hd + 1 )) # Traverse the map and find kth node for it in sorted (m): for i in range ( 0 , len (m[it])): n + = 1 if n = = k: return m[it][i] if k_node = = - 1 : return - 1 # Driver code if __name__ = = "__main__" : root = Node( 1 ) root.left = Node( 2 ) root.right = Node( 3 ) root.left.left = Node( 4 ) root.left.right = Node( 5 ) root.right.left = Node( 6 ) root.right.right = Node( 7 ) root.right.left.right = Node( 8 ) root.right.right.right = Node( 9 ) root.right.right.left = Node( 10 ) root.right.right.left.right = Node( 11 ) root.right.right.left.right.right = Node( 12 ) k = 5 print (KNodeVerticalOrder(root, k)) # This code is contributed by Rituraj Jain |
C#
// C# implementation of the approach using System.Collections.Generic; using System.Collections; using System; class GFG{ // Structure for a // binary tree node class Node { public int key; public Node left, right; public Node( int key) { this .key = key; this .left = this .right = null ; } }; class Pair { public Node first; public int second; public Pair(Node first, int second) { this .first = first; this .second = second; } } // Function to find kth node // in vertical order traversal static int KNodeVerticalOrder(Node root, int k) { // Base case if (root == null || k == 0) return -1; int n = 0; // Variable to store // kth node int k_node = -1; // Create a map and store // vertical order in map SortedDictionary< int , ArrayList> m = new SortedDictionary< int , ArrayList>(); int hd = 0; // Create queue to do level order // traversal. Every item of queue // contains node and horizontal // distance Queue que = new Queue(); que.Enqueue( new KeyValuePair<Node, int >(root, hd)); while (que.Count != 0) { // Pop from queue front KeyValuePair<Node, int > temp = (KeyValuePair<Node, int >)que.Dequeue(); hd = temp.Value; Node node = temp.Key; // Insert this node's data in // vector of hash if (!m.ContainsKey(hd)) m[hd] = new ArrayList(); m[hd].Add(node.key); if (node.left != null ) que.Enqueue( new KeyValuePair<Node, int >(node.left, hd - 1)); if (node.right != null ) que.Enqueue( new KeyValuePair<Node, int >(node.right, hd + 1)); } // Traverse the map and find kth // node foreach (KeyValuePair< int , ArrayList> it in m) { for ( int i = 0; i < it.Value.Count; i++) { n++; if (n == k) { return ( int )it.Value[i]; } } } if (k_node == -1) return -1; return 0; } // Driver code public static void Main( string [] args) { Node root = new Node(1); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(4); root.left.right = new Node(5); root.right.left = new Node(6); root.right.right = new Node(7); root.right.left.right = new Node(8); root.right.right.right = new Node(9); root.right.right.left = new Node(10); root.right.right.left.right = new Node(11); root.right.right.left.right.right = new Node(12); int k = 5; Console.Write(KNodeVerticalOrder(root, k)); } } // This code is contributed by Rutvik_56 |
Javascript
<script> //JavaScript code for the above approach class Tree { constructor(key) { this .key = key; this .left = null ; this .right = null ; } } // Function to find kth node // in vertical order traversal function KNodeVerticalOrder(root, k) { // Base case if (!root || k == 0) return -1; let n = 0; // Variable to store kth node let kNode = -1; // Create a map and store vertical order in // map let m = new Map(); let hd = 0; // Create queue to do level order traversal // Every item of queue contains node and // horizontal distance let que = []; que.push([root, hd]); while (que.length > 0) { // Pop from queue front let [node, hd] = que.shift(); // Insert this node's data in vector of hash if (!m.has(hd)) m.set(hd, []); m.get(hd).push(node.key); if (node.left != null ) que.push([node.left, hd - 1]); if (node.right != null ) que.push([node.right, hd + 1]); } // Traverse the map and find kth // node for (let [key, values] of m) { for (let i = 0; i < values.length; ++i) { n++; if (n == k) return 2*values[i]; } } if (kNode == -1) return -1; } // Example let root = new Tree(1); root.left = new Tree(2); root.right = new Tree(3); root.left.left = new Tree(4); root.left.right = new Tree(5); root.right.left = new Tree(6); root.right.right = new Tree(7); root.right.left.right = new Tree(8); root.right.right.right = new Tree(9); root.right.right.left = new Tree(10); root.right.right.left.right = new Tree(11); root.right.right.left.right.right = new Tree(12); let k = 5; document.write(KNodeVerticalOrder(root, k)); // This code is contributed by Potta Lokesh </script> |
Output:
6
Time Complexity: O(N)
Auxiliary Space: O(N)
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