Find Leftmost and Rightmost node of BST from its given preorder traversal
Given a preorder sequence of the binary search tree of N nodes. The task is to find its leftmost and rightmost node.
Examples:
Input : N = 5, preorder[]={ 3, 2, 1, 5, 4 } Output : Leftmost = 1, Rightmost = 5 The BST constructed from this preorder sequence would be: 3 / \ 2 5 / / 1 4 Leftmost Node of this tree is equal to 1 Rightmost Node of this tree is equal to 5 Input : N = 3 preorder[]={ 2, 1, 3} Output : Leftmost = 1, Rightmost = 3
Naive Approach:
Construct BST from the given preorder sequence.See this post for understanding code to construct bst from given preorder. After constructing the BST, Find leftmost and rightmost node by traversing from root to leftmost node and traversing from root to rightmost node.
Time Complexity: O(N)
Space Complexity: O(N)
Efficient Approach:
Instead of constructing the tree, make use of the property of BST. The leftmost node in BST always has the smallest value and rightmost node in BST always has the largest value.
So from the given array we just need to find minimum value for leftmost node and maximum value for rightmost node.
Below is the implementation of the above approach
C++
// C++ program to find leftmost and // rightmost node from given preorder sequence #include <bits/stdc++.h> using namespace std; // Function to return the leftmost and // rightmost nodes of the BST whose // preorder traversal is given void LeftRightNode( int preorder[], int n) { // Variables for finding minimum // and maximum values of the array int min = INT_MAX, max = INT_MIN; for ( int i = 0; i < n; i++) { // Update the minimum if (min > preorder[i]) min = preorder[i]; // Update the maximum if (max < preorder[i]) max = preorder[i]; } // Print the values cout << "Leftmost node is " << min << "\n" ; cout << "Rightmost node is " << max; } // Driver Code int main() { int preorder[] = { 3, 2, 1, 5, 4 }; int n = 5; LeftRightNode(preorder, n); return 0; } |
Java
// Java program to find leftmost and // rightmost node from given preorder sequence class GFG { // Function to return the leftmost and // rightmost nodes of the BST whose // preorder traversal is given static void LeftRightNode( int preorder[], int n) { // Variables for finding minimum // and maximum values of the array int min = Integer.MAX_VALUE, max = Integer.MIN_VALUE; for ( int i = 0 ; i < n; i++) { // Update the minimum if (min > preorder[i]) min = preorder[i]; // Update the maximum if (max < preorder[i]) max = preorder[i]; } // Print the values System.out.println( "Leftmost node is " + min); System.out.println( "Rightmost node is " + max); } // Driver Code public static void main(String[] args) { int preorder[] = { 3 , 2 , 1 , 5 , 4 }; int n = 5 ; LeftRightNode(preorder, n); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 program to find leftmost and # rightmost node from given preorder sequence # Function to return the leftmost and # rightmost nodes of the BST whose # preorder traversal is given def LeftRightNode(preorder, n): # Variables for finding minimum # and maximum values of the array min = 10 * * 9 max = - 10 * * 9 for i in range (n): # Update the minimum if ( min > preorder[i]): min = preorder[i] # Update the maximum if ( max < preorder[i]): max = preorder[i] # Print the values print ( "Leftmost node is " , min ) print ( "Rightmost node is " , max ) # Driver Code preorder = [ 3 , 2 , 1 , 5 , 4 ] n = len (preorder) LeftRightNode(preorder, n) # This code is contributed by mohit kumar 29 |
C#
// C# program to find leftmost and // rightmost node from given preorder sequence using System; class GFG { // Function to return the leftmost and // rightmost nodes of the BST whose // preorder traversal is given static void LeftRightNode( int []preorder, int n) { // Variables for finding minimum // and maximum values of the array int min = int .MaxValue, max = int .MinValue; for ( int i = 0; i < n; i++) { // Update the minimum if (min > preorder[i]) min = preorder[i]; // Update the maximum if (max < preorder[i]) max = preorder[i]; } // Print the values Console.WriteLine( "Leftmost node is " + min); Console.WriteLine( "Rightmost node is " + max); } // Driver Code public static void Main(String[] args) { int []preorder = { 3, 2, 1, 5, 4 }; int n = 5; LeftRightNode(preorder, n); } } // This code is contributed by Rajput-Ji |
Leftmost node is 1 Rightmost node is 5
Time Complexity: O(N)
Space Complexity: O(1)
Recommended Posts:
- Find n-th node in Preorder traversal of a Binary Tree
- Find postorder traversal of BST from preorder traversal
- Print leftmost and rightmost nodes of a Binary Tree
- Construct Full Binary Tree using its Preorder traversal and Preorder traversal of its mirror tree
- Leftmost and rightmost indices of the maximum and the minimum element of an array
- Find n-th node of inorder traversal
- Find the Kth node in the DFS traversal of a given subtree in a Tree
- Find n-th node in Postorder traversal of a Binary Tree
- Print all the nodes except the leftmost node in every level of the given binary tree
- Find the kth node in vertical order traversal of a Binary Tree
- Morris traversal for Preorder
- Construct BST from given preorder traversal | Set 2
- Iterative Preorder Traversal
- Preorder Traversal of N-ary Tree Without Recursion
- Iterative Preorder Traversal of an N-ary Tree
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.