Second largest element in BST
Given a Binary Search Tree (BST), find the second largest element.
Example:
Input: Root of below BST
10
/
5Output: 5
Input: Root of below BST
10
/ \
5 20
\
30
Output: 20
Source: Microsoft Interview
The idea is similar to below post.
K’th Largest Element in BST when modification to BST is not allowed
The second largest element is second last element in inorder traversal and second element in reverse inorder traversal. We traverse given Binary Search Tree in reverse inorder and keep track of counts of nodes visited. Once the count becomes 2, we print the node.
Below is the implementation of above idea.
C++
// C++ program to find 2nd largest element in BST #include<bits/stdc++.h> using namespace std; struct Node { int key; Node *left, *right; }; // A utility function to create a new BST node Node *newNode( int item) { Node *temp = new Node; temp->key = item; temp->left = temp->right = NULL; return temp; } // A function to find 2nd largest element in a given tree. void secondLargestUtil(Node *root, int &c) { // Base cases, the second condition is important to // avoid unnecessary recursive calls if (root == NULL || c >= 2) return ; // Follow reverse inorder traversal so that the // largest element is visited first secondLargestUtil(root->right, c); // Increment count of visited nodes c++; // If c becomes k now, then this is the 2nd largest if (c == 2) { cout << "2nd largest element is " << root->key << endl; return ; } // Recur for left subtree secondLargestUtil(root->left, c); } // Function to find 2nd largest element void secondLargest(Node *root) { // Initialize count of nodes visited as 0 int c = 0; // Note that c is passed by reference secondLargestUtil(root, c); } /* A utility function to insert a new node with given key in BST */ Node* insert(Node* node, int key) { /* If the tree is empty, return a new node */ if (node == NULL) return newNode(key); /* Otherwise, recur down the tree */ if (key < node->key) node->left = insert(node->left, key); else if (key > node->key) node->right = insert(node->right, key); /* return the (unchanged) node pointer */ return node; } // Driver Program to test above functions int main() { /* Let us create following BST 50 / \ 30 70 / \ / \ 20 40 60 80 */ Node *root = NULL; root = insert(root, 50); insert(root, 30); insert(root, 20); insert(root, 40); insert(root, 70); insert(root, 60); insert(root, 80); secondLargest(root); return 0; } |
Java
class Node { int key; Node left, right; Node( int item) { key = item; left = right = null ; } } public class Main { // A function to find 2nd largest element in a given tree. static void secondLargestUtil(Node root, int c) { // Base cases, the second condition is important to // avoid unnecessary recursive calls if (root == null || c >= 2 ) return ; // Follow reverse inorder traversal so that the // largest element is visited first secondLargestUtil(root.right, c); // Increment count of visited nodes c++; // If c becomes k now, then this is the 2nd largest if (c == 2 ) { System.out.println( "2nd largest element is " + root.key); return ; } // Recur for left subtree secondLargestUtil(root.left, c); } // Function to find 2nd largest element static void secondLargest(Node root) { // Initialize count of nodes visited as 0 int c = 0 ; // Note that c is passed by reference secondLargestUtil(root, c); } // A utility function to insert a new node with given key in BST static Node insert(Node node, int key) { /* If the tree is empty, return a new node */ if (node == null ) return new Node(key); /* Otherwise, recur down the tree */ if (key < node.key) node.left = insert(node.left, key); else if (key > node.key) node.right = insert(node.right, key); /* return the (unchanged) node pointer */ return node; } // Driver Program to test above functions public static void main(String[] args) { /* Let us create following BST 50 / \ 30 70 / \ / \ 20 40 60 80 */ Node root = null ; root = insert(root, 50 ); insert(root, 30 ); insert(root, 20 ); insert(root, 40 ); insert(root, 70 ); insert(root, 60 ); insert(root, 80 ); secondLargest(root); } } |
Python3
# Python3 code to find second largest # element in BST class Node: # Constructor to create a new node def __init__( self , data): self .key = data self .left = None self .right = None # A function to find 2nd largest # element in a given tree. def secondLargestUtil(root, c): # Base cases, the second condition # is important to avoid unnecessary # recursive calls if root = = None or c[ 0 ] > = 2 : return # Follow reverse inorder traversal so that # the largest element is visited first secondLargestUtil(root.right, c) # Increment count of visited nodes c[ 0 ] + = 1 # If c becomes k now, then this is # the 2nd largest if c[ 0 ] = = 2 : print ( "2nd largest element is" , root.key) return # Recur for left subtree secondLargestUtil(root.left, c) # Function to find 2nd largest element def secondLargest(root): # Initialize count of nodes # visited as 0 c = [ 0 ] # Note that c is passed by reference secondLargestUtil(root, c) # A utility function to insert a new # node with given key in BST def insert(node, key): # If the tree is empty, return a new node if node = = None : return Node(key) # Otherwise, recur down the tree if key < node.key: node.left = insert(node.left, key) elif key > node.key: node.right = insert(node.right, key) # return the (unchanged) node pointer return node # Driver Code if __name__ = = '__main__' : # Let us create following BST # 50 # / \ # 30 70 # / \ / \ # 20 40 60 80 root = None root = insert(root, 50 ) insert(root, 30 ) insert(root, 20 ) insert(root, 40 ) insert(root, 70 ) insert(root, 60 ) insert(root, 80 ) secondLargest(root) # This code is contributed by PranchalK |
C#
using System; // C# code to find the second largest element in a BST // A binary tree node public class Node { public int Data { get ; set ; } public Node Left { get ; set ; } public Node Right { get ; set ; } public Node( int data) { Data = data; Left = Right = null ; } } public class BinarySearchTree { // Root of the BST public Node Root { get ; set ; } // Constructor public BinarySearchTree() { Root = null ; } // Function to insert new nodes public void Insert( int data) { Root = InsertNode(Root, data); } /* A utility function to insert a new node with a given key in BST */ private Node InsertNode(Node node, int data) { /* If the tree is empty, return a new node */ if (node == null ) { Root = new Node(data); return Root; } /* Otherwise, recur down the tree */ if (data < node.Data) node.Left = InsertNode(node.Left, data); else node.Right = InsertNode(node.Right, data); return node; } // class that stores the value of count private class Counter { private readonly BinarySearchTree outerInstance; public Counter(BinarySearchTree outerInstance) { this .outerInstance = outerInstance; } public int Count { get ; set ; } } // Function to find the 2nd largest element private void FindSecondLargestElement(Node node, Counter counter) { // Base cases, the second condition is important to // avoid unnecessary recursive calls if (node == null || counter.Count >= 2) return ; // Follow reverse inorder traversal so that the // The largest element is visited first FindSecondLargestElement(node.Right, counter); // Increment count of visited nodes counter.Count++; // If counter becomes k now, then this is the 2nd largest if (counter.Count == 2) { Console.Write( "The 2nd largest element is " + node.Data); return ; } // Recur for Left subtree FindSecondLargestElement(node.Left, counter); } // Function to find 2nd largest element public void GetSecondLargestNode(Node node) { // object of counter class Counter counter = new Counter( this ); FindSecondLargestElement(Root, counter); } // Driver function public static void Main( string [] args) { BinarySearchTree bst = new BinarySearchTree(); /* Let us create the following BST 50 / \ 30 70 / \ / \ 20 40 60 80 */ bst.Insert(50); bst.Insert(30); bst.Insert(20); bst.Insert(40); bst.Insert(70); bst.Insert(60); bst.Insert(80); bst.GetSecondLargestNode(bst.Root); } } // This code is contributed by Shrikant13 // This class is edited by Alireza Maleki |
Javascript
<script> // JavaScript code to find second largest // element in BST // A binary tree node class Node { constructor(d) { this .data = d; this .left = this .right = null ; } } // Root of BST var root = null ; // function to insert new nodes function insert(data) { this .root = this .insertRec( this .root, data); } /* A utility function to insert a new node with given key in BST */ function insertRec(node , data) { /* If the tree is empty, return a new node */ if (node == null ) { this .root = new Node(data); return this .root; } /* Otherwise, recur down the tree */ if (data < node.data) { node.left = this .insertRec(node.left, data); } else { node.right = this .insertRec(node.right, data); } return node; } // class that stores the value of count class count { constructor(){ this .c = 0; } } // Function to find 2nd largest element function secondLargestUtil(node, C) { // Base cases, the second condition is important to // avoid unnecessary recursive calls if (node == null || C.c >= 2) return ; // Follow reverse inorder traversal so that the // largest element is visited first this .secondLargestUtil(node.right, C); // Increment count of visited nodes C.c++; // If c becomes k now, then this is the 2nd largest if (C.c == 2) { document.write( "2nd largest element is " + node.data); return ; } // Recur for left subtree this .secondLargestUtil(node.left, C); } // Function to find 2nd largest element function secondLargest(node) { // object of class count var C = new count(); this .secondLargestUtil( this .root, C); } // Driver function /* Let us create following BST 50 / \ 30 70 / \ / \ 20 40 60 80 */ insert(50); insert(30); insert(20); insert(40); insert(70); insert(60); insert(80); secondLargest(root); // This code contributed by aashish1995 </script> |
Output
2nd largest element is 70
Time complexity : O(h) where h is height of BST.
Space Complexity: O(h) for call stack where h is height of BST
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