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; } |
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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 |
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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 |
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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 |
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Output:
Leftmost node is 1 Rightmost node is 5
Time Complexity: O(N)
Space Complexity: O(1)
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