Print extreme nodes of each level of Binary Tree in alternate order
Given a binary tree, print nodes of extreme corners of each level but in alternate order.
Example:
For above tree, the output can be
1 2 7 8 31
– print rightmost node of 1st level
– print leftmost node of 2nd level
– print rightmost node of 3rd level
– print leftmost node of 4th level
– print rightmost node of 5th level
OR
1 3 4 15 16
– print leftmost node of 1st level
– print rightmost node of 2nd level
– print leftmost node of 3rd level
– print rightmost node of 4th level
– print leftmost node of 5th level
The idea is to traverse tree level by level. For each level, we count number of nodes in it and print its leftmost or the rightmost node based on value of a Boolean flag. We dequeue all nodes of current level and enqueue all nodes of next level and invert value of Boolean flag when switching levels.
Below is the implementation of above idea –
C++
/* C++ program to print nodes of extreme corners of each level in alternate order */#include <bits/stdc++.h> using namespace std; /* A binary tree node has data, pointer to left child and a pointer to right child */struct Node { int data; Node *left, *right; }; /* Helper function that allocates a new node with the given data and NULL left and right pointers. */Node* newNode(int data) { Node* node = new Node; node->data = data; node->right = node->left = NULL; return node; } /* Function to print nodes of extreme corners of each level in alternate order */void printExtremeNodes(Node* root) { if (root == NULL) return; // Create a queue and enqueue left and right // children of root queue<Node*> q; q.push(root); // flag to indicate whether leftmost node or // the rightmost node has to be printed bool flag = false; while (!q.empty()) { // nodeCount indicates number of nodes // at current level. int nodeCount = q.size(); int n = nodeCount; // Dequeue all nodes of current level // and Enqueue all nodes of next level while (n--) { Node* curr = q.front(); // Enqueue left child if (curr->left) q.push(curr->left); // Enqueue right child if (curr->right) q.push(curr->right); // Dequeue node q.pop(); // if flag is true, print leftmost node if (flag && n == nodeCount - 1) cout << curr->data << " "; // if flag is false, print rightmost node if (!flag && n == 0) cout << curr->data << " "; } // invert flag for next level flag = !flag; } } /* Driver program to test above functions */int main() { // Binary Tree of Height 4 Node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); root->right->right = newNode(7); root->left->left->left = newNode(8); root->left->left->right = newNode(9); root->left->right->left = newNode(10); root->left->right->right = newNode(11); root->right->right->left = newNode(14); root->right->right->right = newNode(15); root->left->left->left->left = newNode(16); root->left->left->left->right = newNode(17); root->right->right->right->right = newNode(31); printExtremeNodes(root); return 0; } |
Java
// Java program to print nodes of extreme corners //of each level in alternate order import java.util.*; class GFG { // A binary tree node has data, pointer to left child //and a pointer to right child / static class Node { int data; Node left, right; }; // Helper function that allocates a new node with the //given data and null left and right pointers. / static Node newNode(int data) { Node node = new Node(); node.data = data; node.right = node.left = null; return node; } // Function to print nodes of extreme corners //of each level in alternate order static void printExtremeNodes(Node root) { if (root == null) return; // Create a queue and enqueue left and right // children of root Queue<Node> q = new LinkedList<Node>(); q.add(root); // flag to indicate whether leftmost node or // the rightmost node has to be printed boolean flag = false; while (q.size()>0) { // nodeCount indicates number of nodes // at current level. int nodeCount = q.size(); int n = nodeCount; // Dequeue all nodes of current level // and Enqueue all nodes of next level while (n-->0) { Node curr = q.peek(); // Enqueue left child if (curr.left!=null) q.add(curr.left); // Enqueue right child if (curr.right!=null) q.add(curr.right); // Dequeue node q.remove(); // if flag is true, print leftmost node if (flag && n == nodeCount - 1) System.out.print( curr.data + " "); // if flag is false, print rightmost node if (!flag && n == 0) System.out.print( curr.data + " "); } // invert flag for next level flag = !flag; } } // Driver code public static void main(String args[]) { // Binary Tree of Height 4 Node root = newNode(1); root.left = newNode(2); root.right = newNode(3); root.left.left = newNode(4); root.left.right = newNode(5); root.right.right = newNode(7); root.left.left.left = newNode(8); root.left.left.right = newNode(9); root.left.right.left = newNode(10); root.left.right.right = newNode(11); root.right.right.left = newNode(14); root.right.right.right = newNode(15); root.left.left.left.left = newNode(16); root.left.left.left.right = newNode(17); root.right.right.right.right = newNode(31); printExtremeNodes(root); } } // This code is contributed by Arnab Kundu |
Python
# Python program to print nodes of extreme corners # of each level in alternate order # Utility class to create a node class Node: def __init__(self, key): self.data = key self.left = self.right = None # Utility function to create a tree node def newNode( data): temp = Node(0) temp.data = data temp.left = temp.right = None return temp # Function to print nodes of extreme corners # of each level in alternate order def printExtremeNodes( root): if (root == None): return # Create a queue and enqueue left and right # children of root q = [] q.append(root) # flag to indicate whether leftmost node or # the rightmost node has to be printed flag = False while (len(q) > 0): # nodeCount indicates number of nodes # at current level. nodeCount = len(q) n = nodeCount # Dequeue all nodes of current level # and Enqueue all nodes of next level while (n > 0): n = n - 1 curr = q[0] # Enqueue left child if (curr.left != None): q.append(curr.left) # Enqueue right child if (curr.right != None): q.append(curr.right) # Dequeue node q.pop(0) # if flag is true, print leftmost node if (flag and n == nodeCount - 1): print( curr.data , end=" ") # if flag is false, print rightmost node if (not flag and n == 0): print( curr.data ,end= " ") # invert flag for next level flag = not flag # Driver program to test above functions # Binary Tree of Height 4 root = newNode(1) root.left = newNode(2) root.right = newNode(3) root.left.left = newNode(4) root.left.right = newNode(5) root.right.right = newNode(7) root.left.left.left = newNode(8) root.left.left.right = newNode(9) root.left.right.left = newNode(10) root.left.right.right = newNode(11) root.right.right.left = newNode(14) root.right.right.right = newNode(15) root.left.left.left.left = newNode(16) root.left.left.left.right = newNode(17) root.right.right.right.right = newNode(31) printExtremeNodes(root) # This code is contributed by Arnab Kundu |
C#
// C# program to print nodes of extreme corners //of each level in alternate order using System; using System.Collections.Generic; class GFG { // A binary tree node has data, pointer to left child //and a pointer to right child / public class Node { public int data; public Node left, right; }; // Helper function that allocates a new node with the //given data and null left and right pointers. / static Node newNode(int data) { Node node = new Node(); node.data = data; node.right = node.left = null; return node; } // Function to print nodes of extreme corners //of each level in alternate order static void printExtremeNodes(Node root) { if (root == null) return; // Create a queue and enqueue left and right // children of root Queue<Node> q = new Queue<Node>(); q.Enqueue(root); // flag to indicate whether leftmost node or // the rightmost node has to be printed Boolean flag = false; while (q.Count > 0) { // nodeCount indicates number of nodes // at current level. int nodeCount = q.Count; int n = nodeCount; // Dequeue all nodes of current level // and Enqueue all nodes of next level while (n-->0) { Node curr = q.Peek(); // Enqueue left child if (curr.left != null) q.Enqueue(curr.left); // Enqueue right child if (curr.right != null) q.Enqueue(curr.right); // Dequeue node q.Dequeue(); // if flag is true, print leftmost node if (flag && n == nodeCount - 1) Console.Write( curr.data + " "); // if flag is false, print rightmost node if (!flag && n == 0) Console.Write( curr.data + " "); } // invert flag for next level flag = !flag; } } // Driver code public static void Main(String []args) { // Binary Tree of Height 4 Node root = newNode(1); root.left = newNode(2); root.right = newNode(3); root.left.left = newNode(4); root.left.right = newNode(5); root.right.right = newNode(7); root.left.left.left = newNode(8); root.left.left.right = newNode(9); root.left.right.left = newNode(10); root.left.right.right = newNode(11); root.right.right.left = newNode(14); root.right.right.right = newNode(15); root.left.left.left.left = newNode(16); root.left.left.left.right = newNode(17); root.right.right.right.right = newNode(31); printExtremeNodes(root); } } // This code is contributed by Rajput-Ji |
Output:
1 2 7 8 31
Time complexity of above solution is O(n) where n is total number of nodes in given binary tree.
Exercise – Print nodes of extreme corners of each level from bottom to top in alternate order.
This article is contributed by Aditya Goel. 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 write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
