Second largest element in BST
Given a Binary Search Tree(BST), find the second largest element.
Examples:
Input: Root of below BST
10
/
5
Output: 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 nodeNode *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 elementvoid 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 functionsint 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
// Java code to find second largest element in BST // A binary tree nodeclass Node { int data; Node left, right; Node(int d) { data = d; left = right = null; }} class BinarySearchTree { // Root of BST Node root; // Constructor BinarySearchTree() { root = null; } // function to insert new nodes public void insert(int data) { this.root = this.insertRec(this.root, data); } /* A utility function to insert a new node with given key in BST */ Node insertRec(Node node, int 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 public class count { int c = 0; } // Function to find 2nd largest element void secondLargestUtil(Node node, count 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) { System.out.print("2nd largest element is "+ node.data); return; } // Recur for left subtree this.secondLargestUtil(node.left, C); } // Function to find 2nd largest element void secondLargest(Node node) { // object of class count count C = new count(); this.secondLargestUtil(this.root, C); } // Driver function public static void main(String[] args) { BinarySearchTree tree = new BinarySearchTree(); /* Let us create following BST 50 / \ 30 70 / \ / \ 20 40 60 80 */ tree.insert(50); tree.insert(30); tree.insert(20); tree.insert(40); tree.insert(70); tree.insert(60); tree.insert(80); tree.secondLargest(tree.root); }} // This code is contributed by Kamal Rawal |
Python3
# Python3 code to find second largest # element in BSTclass 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 Codeif __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 second largest element in BST // A binary tree node public class Node{ public int data; public Node left, right; public Node(int d) { data = d; left = right = null; }} public class BinarySearchTree{ // Root of BST public Node root; // Constructor public BinarySearchTree() { root = null; } // function to insert new nodes public virtual void insert(int data) { this.root = this.insertRec(this.root, data); } /* A utility function to insert a new node with given key in BST */ public virtual Node insertRec(Node node, int 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 public class count { private readonly BinarySearchTree outerInstance; public count(BinarySearchTree outerInstance) { this.outerInstance = outerInstance; } public int c = 0; } // Function to find 2nd largest element public virtual void secondLargestUtil(Node node, count 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) { Console.Write("2nd largest element is " + node.data); return; } // Recur for left subtree this.secondLargestUtil(node.left, C); } // Function to find 2nd largest element public virtual void secondLargest(Node node) { // object of class count count C = new count(this); this.secondLargestUtil(this.root, C); } // Driver function public static void Main(string[] args) { BinarySearchTree tree = new BinarySearchTree(); /* Let us create following BST 50 / \ 30 70 / \ / \ 20 40 60 80 */ tree.insert(50); tree.insert(30); tree.insert(20); tree.insert(40); tree.insert(70); tree.insert(60); tree.insert(80); tree.secondLargest(tree.root); }} // This code is contributed by Shrikant13 |
Output:
2nd largest element is 70
Time complexity of the above solution is O(h) where h is height of BST.
This article is contributed by Ravi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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