Recursive Program to Print extreme nodes of each level of Binary Tree in alternate order
Given a binary tree, the task is to print nodes of extreme corners of each level but in alternate order.
Examples:
Input : 1 / \ 2 3 / / \ 4 5 6 / / \ 7 8 9 Output : 1 2 6 7 Print the rightmost node at 1st level: 1 Print the leftmost node at 2nd level: 2 Print the rightmost node at 3rd level: 6 Print the leftmost node at 4th level: 7 Other possible output will be -> 1 3 4 9 Input : 3 / \ 8 1 / \ / \ 9 5 6 4 Output : 3 8 4
We have already discussed the iterative approach to solve this problem. In this post the recursive approach is discussed.
Approach: The idea is to perform level order traversal in the spiral form and at each level print the first node during the traversal, these will be the nodes at extreme corner present in the alternate form.
Below is the implementation of the above approach:
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 struct Node { int data; Node *left, *right; }; // Utility function to allocate memory for a new node Node* newNode( int data) { Node* node = new (Node); node->data = data; node->left = node->right = NULL; return (node); } // Function that returns the height of the binary tree int height(Node* root) { if (root == NULL) return 0; int lheight = height(root->left); int rheight = height(root->right); return max(lheight, rheight) + 1; } // Function performs level order traversal from right to // left and prints the first node during the traversal void rightToLeft(Node* root, int level, int & f) { if (root == NULL) return ; // Checks for the value of f so that // only first node is printed during // the traversal and no other node is printed if (level == 1 && f == 0) { printf ( "%d " , root->data); f = 1; } else if (level > 1) { rightToLeft(root->right, level - 1, f); rightToLeft(root->left, level - 1, f); } } // Function performs level order traversal from left to // right and prints the first node during the traversal void leftToRight(Node* root, int level, int & f) { if (root == NULL) return ; // Checks for the value of f so that // only first node is printed during // the traversal and no other node is printed if (level == 1 && f == 1) { printf ( "%d " , root->data); f = 0; } else if (level > 1) { leftToRight(root->left, level - 1, f); leftToRight(root->right, level - 1, f); } } // Function to print the extreme nodes of // a given binary tree void printExtremeNodes(Node* root) { // Stores height of binary tree int h = height(root); // Flag to mark the change in level int flag = 0; // To check if the extreme node of a // particular level has been visited int f = 0; for ( int i = 1; i <= h; i++) { // If flag is zero then traverse from // right to left at the given level and // print the first node during the traversal if (flag == 0) { rightToLeft(root, i, f); flag = 1; } // If flag is one then traverse from // left to right at the given level and // print the first node during the traversal else if (flag == 1) { leftToRight(root, i, f); flag = 0; } } return ; } // Driver code int main() { 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 { //INT class static class INT { int a; } // A binary tree node static class Node { int data; Node left, right; }; // Utility function to allocate memory for a new node static Node newNode( int data) { Node node = new Node(); node.data = data; node.left = node.right = null ; return (node); } // Function that returns the height of the binary tree static int height(Node root) { if (root == null ) return 0 ; int lheight = height(root.left); int rheight = height(root.right); return Math.max(lheight, rheight) + 1 ; } // Function performs level order traversal from right to // left and prints the first node during the traversal static void rightToLeft(Node root, int level, INT f) { if (root == null ) return ; // Checks for the value of f so that // only first node is printed during // the traversal and no other node is printed if (level == 1 && f.a == 0 ) { System.out.printf( "%d " , root.data); f.a = 1 ; } else if (level > 1 ) { rightToLeft(root.right, level - 1 , f); rightToLeft(root.left, level - 1 , f); } } // Function performs level order traversal from left to // right and prints the first node during the traversal static void leftToRight(Node root, int level, INT f) { if (root == null ) return ; // Checks for the value of f so that // only first node is printed during // the traversal and no other node is printed if (level == 1 && f.a == 1 ) { System.out.printf( "%d " , root.data); f.a = 0 ; } else if (level > 1 ) { leftToRight(root.left, level - 1 , f); leftToRight(root.right, level - 1 , f); } } // Function to print the extreme nodes of // a given binary tree static void printExtremeNodes(Node root) { // Stores height of binary tree int h = height(root); // Flag to mark the change in level int flag = 0 ; // To check if the extreme node of a // particular level has been visited INT f= new INT(); f.a = 0 ; for ( int i = 1 ; i <= h; i++) { // If flag is zero then traverse from // right to left at the given level and // print the first node during the traversal if (flag == 0 ) { rightToLeft(root, i, f); flag = 1 ; } // If flag is one then traverse from // left to right at the given level and // print the first node during the traversal else if (flag == 1 ) { leftToRight(root, i, f); flag = 0 ; } } return ; } // Driver code public static void main(String args[]) { 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 |
Python3
# Python3 program to print nodes of # extreme corners of each level in # alternate order from collections import deque # A binary tree node has key, pointer to left # child and a pointer to right child class Node: def __init__( self , key): self .data = key self .left = None self .right = None # Function that returns the height of # the binary tree def height(root: Node) - > int : if (root is None ): return 0 lheight = height(root.left) rheight = height(root.right) return max (lheight, rheight) + 1 # Function performs level order traversal # from right to left and prints the first # node during the traversal def rightToLeft(root: Node, level: int ) - > int : global f if (root is None ): return # Checks for the value of f so that # only first node is printed during # the traversal and no other node is printed if (level = = 1 and f = = 0 ): print ( "%d " % root.data, end = "") f = 1 elif (level > 1 ): rightToLeft(root.right, level - 1 ) rightToLeft(root.left, level - 1 ) # Function performs level order traversal # from left to right and prints the first # node during the traversal def leftToRight(root: Node, level: int ): global f if (root is None ): return # Checks for the value of f so that # only first node is printed during # the traversal and no other node is printed if (level = = 1 and f = = 1 ): print ( "%d " % root.data, end = "") f = 0 elif (level > 1 ): leftToRight(root.left, level - 1 ) leftToRight(root.right, level - 1 ) # Function to print the extreme nodes of # a given binary tree def printExtremeNodes(root: Node): global f # Stores height of binary tree h = height(root) # Flag to mark the change in level flag = 0 # To check if the extreme node of a # particular level has been visited f = 0 for i in range ( 1 , h + 1 ): # If flag is zero then traverse from # right to left at the given level and # print the first node during the traversal if (flag = = 0 ): rightToLeft(root, i) flag = 1 # If flag is one then traverse from # left to right at the given level and # print the first node during the traversal elif (flag = = 1 ): leftToRight(root, i) flag = 0 return # Driver Code if __name__ = = "__main__" : root = Node( 1 ) root.left = Node( 2 ) root.right = Node( 3 ) root.left.left = Node( 4 ) root.left.right = Node( 5 ) root.right.right = Node( 7 ) root.left.left.left = Node( 8 ) root.left.left.right = Node( 9 ) root.left.right.left = Node( 10 ) root.left.right.right = Node( 11 ) root.right.right.left = Node( 14 ) root.right.right.right = Node( 15 ) root.left.left.left.left = Node( 16 ) root.left.left.left.right = Node( 17 ) root.right.right.right.right = Node( 31 ) printExtremeNodes(root) # This code is contributed by sanjeev2552 |
C#
// C# program to print nodes of extreme corners // of each level in alternate order using System; class GFG { //INT class public class INT { public int a; } // A binary tree node public class Node { public int data; public Node left, right; }; // Utility function to allocate memory for a new node static Node newNode( int data) { Node node = new Node(); node.data = data; node.left = node.right = null ; return (node); } // Function that returns the height of the binary tree static int height(Node root) { if (root == null ) return 0; int lheight = height(root.left); int rheight = height(root.right); return Math.Max(lheight, rheight) + 1; } // Function performs level order traversal from right to // left and prints the first node during the traversal static void rightToLeft(Node root, int level, INT f) { if (root == null ) return ; // Checks for the value of f so that // only first node is printed during // the traversal and no other node is printed if (level == 1 && f.a == 0) { Console.Write( "{0} " , root.data); f.a = 1; } else if (level > 1) { rightToLeft(root.right, level - 1, f); rightToLeft(root.left, level - 1, f); } } // Function performs level order traversal from left to // right and prints the first node during the traversal static void leftToRight(Node root, int level, INT f) { if (root == null ) return ; // Checks for the value of f so that // only first node is printed during // the traversal and no other node is printed if (level == 1 && f.a == 1) { Console.Write( "{0} " , root.data); f.a = 0; } else if (level > 1) { leftToRight(root.left, level - 1, f); leftToRight(root.right, level - 1, f); } } // Function to print the extreme nodes of // a given binary tree static void printExtremeNodes(Node root) { // Stores height of binary tree int h = height(root); // Flag to mark the change in level int flag = 0; // To check if the extreme node of a // particular level has been visited INT f= new INT(); f.a = 0; for ( int i = 1; i <= h; i++) { // If flag is zero then traverse from // right to left at the given level and // print the first node during the traversal if (flag == 0) { rightToLeft(root, i, f); flag = 1; } // If flag is one then traverse from // left to right at the given level and // print the first node during the traversal else if (flag == 1) { leftToRight(root, i, f); flag = 0; } } return ; } // Driver code public static void Main() { 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 contributed by PrinciRaj1992 */ |
Javascript
<script> // JavaScript program to print nodes of extreme corners // of each level in alternate order //INT class class INT { constructor() { this .a = 0; } } // A binary tree node class Node { constructor() { this .data = 0; this .left = null ; this .right = null ; } }; // Utility function to allocate memory for a new node function newNode(data) { var node = new Node(); node.data = data; node.left = node.right = null ; return (node); } // Function that returns the height of the binary tree function height(root) { if (root == null ) return 0; var lheight = height(root.left); var rheight = height(root.right); return Math.max(lheight, rheight) + 1; } // Function performs level order traversal from right to // left and prints the first node during the traversal function rightToLeft(root, level, f) { if (root == null ) return ; // Checks for the value of f so that // only first node is printed during // the traversal and no other node is printed if (level == 1 && f.a == 0) { document.write(root.data + " " ); f.a = 1; } else if (level > 1) { rightToLeft(root.right, level - 1, f); rightToLeft(root.left, level - 1, f); } } // Function performs level order traversal from left to // right and prints the first node during the traversal function leftToRight(root, level, f) { if (root == null ) return ; // Checks for the value of f so that // only first node is printed during // the traversal and no other node is printed if (level == 1 && f.a == 1) { document.write(root.data+ " " ); f.a = 0; } else if (level > 1) { leftToRight(root.left, level - 1, f); leftToRight(root.right, level - 1, f); } } // Function to print the extreme nodes of // a given binary tree function printExtremeNodes(root) { // Stores height of binary tree var h = height(root); // Flag to mark the change in level var flag = 0; // To check if the extreme node of a // particular level has been visited var f= new INT(); f.a = 0; for ( var i = 1; i <= h; i++) { // If flag is zero then traverse from // right to left at the given level and // print the first node during the traversal if (flag == 0) { rightToLeft(root, i, f); flag = 1; } // If flag is one then traverse from // left to right at the given level and // print the first node during the traversal else if (flag == 1) { leftToRight(root, i, f); flag = 0; } } return ; } // Driver code var 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); </script> |
Output:
1 2 7 8 31
Time Complexity: O(N^2), where N is the total number of nodes in the binary tree.
Auxiliary Space: O(N)