Reverse Level Order traversal in spiral form
Given a binary tree, the task is to print the reverse level order in spiral form.
Examples:
Input:
1
/ \
2 3
/ \ / \
4 5 6 7
Output: 4 5 6 7 3 2 1
Input:
5
/ \
6 4
/ \ /
7 1 8
\ \
3 10
Output: 3 10 8 1 7 6 4 5
Approach: The idea is to traverse the tree in a Reverse Level Order manner but with a slight modification. We will use a variable flag and initially set it’s value to one. As we complete the reverse level order traversal of the tree, from left to right we will set the value of flag to zero, so that next time we traverse the Tree from right to left and as we complete the traversal we set it’s value back to one. We will repeat this whole step until we have traversed the Binary Tree completely.
Below is the implementation of the above approach:
C++
// C++ program to print reverse level order // traversal of a binary tree in spiral form #include <bits/stdc++.h> using namespace std; // Binary tree Node struct Node { struct Node* left; struct Node* right; int data; Node(int data) { this->data = data; this->left = NULL; this->right = NULL; } }; // Recursive Function to find height // of binary tree int height(struct Node* root) { if (root == NULL) return 0; // Compute the height of each subtree int lheight = height(root->left); int rheight = height(root->right); // Return the maximum of two return max(lheight + 1, rheight + 1); } // Function to Print Nodes from right to left void rightToLeft(struct Node* root, int level) { if (root == NULL) return; if (level == 1) cout << root->data << " "; else if (level > 1) { rightToLeft(root->right, level - 1); rightToLeft(root->left, level - 1); } } // Function to Print Nodes from left to right void leftToRight(struct Node* root, int level) { if (root == NULL) return; if (level == 1) cout << root->data << " "; else if (level > 1) { leftToRight(root->left, level - 1); leftToRight(root->right, level - 1); } } // Function to print Reverse level order traversal // of a Binary tree in spiral form void reverseSpiral(struct Node* root) { // Flag is used to mark the change // in level int flag = 1; // Height of tree int h = height(root); for (int i = h; i >= 1; i--) { // If flag value is one print Nodes // from left to right if (flag == 1) { leftToRight(root, i); // Mark flag as zero so that next time // tree is traversed from right to left flag = 0; } // If flag is zero print Nodes // from right to left else if (flag == 0) { rightToLeft(root, i); // Mark flag as one so that next time // Nodes are printed from left to right flag = 1; } } } // Driver code int main() { struct Node* root = new Node(5); root->left = new Node(9); root->right = new Node(3); root->left->left = new Node(6); root->right->right = new Node(4); root->left->left->left = new Node(8); root->left->left->right = new Node(7); reverseSpiral(root); return 0; } |
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Java
// Java program to print reverse level order // traversal of a binary tree in spiral form class Solution { // Binary tree Node static class Node { Node left; Node right; int data; Node(int data) { this.data = data; this.left = null; this.right = null; } }; // Recursive Function to find height // of binary tree static int height(Node root) { if (root == null) return 0; // Compute the height of each subtree int lheight = height(root.left); int rheight = height(root.right); // Return the maximum of two return Math.max(lheight + 1, rheight + 1); } // Function to Print Nodes from right to left static void rightToLeft(Node root, int level) { if (root == null) return; if (level == 1) System.out.print(root.data + " "); else if (level > 1) { rightToLeft(root.right, level - 1); rightToLeft(root.left, level - 1); } } // Function to Print Nodes from left to right static void leftToRight(Node root, int level) { if (root == null) return; if (level == 1) System.out.print(root.data + " "); else if (level > 1) { leftToRight(root.left, level - 1); leftToRight(root.right, level - 1); } } // Function to print Reverse level order traversal // of a Binary tree in spiral form static void reverseSpiral(Node root) { // Flag is used to mark the change // in level int flag = 1; // Height of tree int h = height(root); for (int i = h; i >= 1; i--) { // If flag value is one print Nodes // from left to right if (flag == 1) { leftToRight(root, i); // Mark flag as zero so that next time // tree is traversed from right to left flag = 0; } // If flag is zero print Nodes // from right to left else if (flag == 0) { rightToLeft(root, i); // Mark flag as one so that next time // Nodes are printed from left to right flag = 1; } } } // Driver code public static void main(String args[]) { Node root = new Node(5); root.left = new Node(9); root.right = new Node(3); root.left.left = new Node(6); root.right.right = new Node(4); root.left.left.left = new Node(8); root.left.left.right = new Node(7); reverseSpiral(root); } } // This code is contributed by Arnab Kundu |
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Python3
# Python3 program to prreverse level order # traversal of a binary tree in spiral form class newNode: # Construct to create a newNode def __init__(self, key): self.data = key self.left = None self.right = None # Recursive Function to find height # of binary tree def height(root): if (root == None): return 0 # Compute the height of each subtree lheight = height(root.left) rheight = height(root.right) # Return the maximum of two return max(lheight + 1, rheight + 1) # Function to PrNodes from right to left def rightToLeft(root, level): if (root == None): return if (level == 1): print(root.data, end =" ") elif (level > 1): rightToLeft(root.right, level - 1) rightToLeft(root.left, level - 1) # Function to PrNodes from left to right def leftToRight(root, level): if (root == None): return if (level == 1) : print(root.data, end = " ") elif (level > 1): leftToRight(root.left, level - 1) leftToRight(root.right, level - 1) # Function to prReverse level order traversal # of a Binary tree in spiral form def reverseSpiral(root): # Flag is used to mark the # change in level flag = 1 # Height of tree h = height(root) for i in range(h, 0, -1): # If flag value is one prNodes # from left to right if (flag == 1): leftToRight(root, i) # Mark flag as zero so that next time # tree is traversed from right to left flag = 0 # If flag is zero prNodes # from right to left elif (flag == 0): rightToLeft(root, i) # Mark flag as one so that next time # Nodes are printed from left to right flag = 1 # Driver Code if __name__ == '__main__': root = newNode(5) root.left = newNode(9) root.right = newNode(3) root.left.left = newNode(6) root.right.right = newNode(4) root.left.left.left = newNode(8) root.left.left.right = newNode(7) reverseSpiral(root) # This code is contributed # by SHUBHAMSINGH10 |
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C#
// C# program to print reverse level order // traversal of a binary tree in spiral form using System; class GFG { // Binary tree Node public class Node { public Node left; public Node right; public int data; public Node(int data) { this.data = data; this.left = null; this.right = null; } }; // Recursive Function to find height // of binary tree static int height(Node root) { if (root == null) return 0; // Compute the height of each subtree int lheight = height(root.left); int rheight = height(root.right); // Return the maximum of two return Math.Max(lheight + 1, rheight + 1); } // Function to Print Nodes from right to left static void rightToLeft(Node root, int level) { if (root == null) return; if (level == 1) Console.Write(root.data + " "); else if (level > 1) { rightToLeft(root.right, level - 1); rightToLeft(root.left, level - 1); } } // Function to Print Nodes from left to right static void leftToRight(Node root, int level) { if (root == null) return; if (level == 1) Console.Write(root.data + " "); else if (level > 1) { leftToRight(root.left, level - 1); leftToRight(root.right, level - 1); } } // Function to print Reverse level order traversal // of a Binary tree in spiral form static void reverseSpiral(Node root) { // Flag is used to mark the change // in level int flag = 1; // Height of tree int h = height(root); for (int i = h; i >= 1; i--) { // If flag value is one print Nodes // from left to right if (flag == 1) { leftToRight(root, i); // Mark flag as zero so that next time // tree is traversed from right to left flag = 0; } // If flag is zero print Nodes // from right to left else if (flag == 0) { rightToLeft(root, i); // Mark flag as one so that next time // Nodes are printed from left to right flag = 1; } } } // Driver code public static void Main(String[] args) { Node root = new Node(5); root.left = new Node(9); root.right = new Node(3); root.left.left = new Node(6); root.right.right = new Node(4); root.left.left.left = new Node(8); root.left.left.right = new Node(7); reverseSpiral(root); } } /* This code is contributed by PrinciRaj1992 */ |
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Output:
8 7 4 6 9 3 5
Time Complexity Analysis : The best case time complexity of reverseSpiral() is O(n+n/2+n/4…) = O(n) in case of a perfect binary tree. And the worst case time complexity is O(n**2) [O((n)+(n-1)+(n-2)+…] in case of a skewed tree taken as input.
Iterative Implementation –
The above problem can be solved with the help of queue and stack. Just perform a normal spiral traversal, use stack to use reverse the elements alternatively. Below is the implementation of the same –
#include <bits/stdc++.h> using namespace std; // Node structure struct Node { int data; Node* left; Node* right; Node(int val) { data = val; left = right = NULL; } }; // function to perform // reversal spiral traversal void reverseSpiral(Node* root) { if (root == NULL) { return; } queue<Node*> q; stack<int> s; bool reverse = true; q.push(root); // iterate until queue is empty while (!q.empty()) { // calculate size of level int count = q.size(); // iterate over level nodes for (int i = 0; i < count; i++) { Node* curr = q.front(); q.pop(); s.push(curr->data); if (reverse) { if (curr->right) { q.push(curr->right); } if (curr->left) { q.push(curr->left); } } else { if (curr->left) { q.push(curr->left); } if (curr->right) { q.push(curr->right); } } } reverse = !reverse; } // pop and print reversed nodes while (!s.empty()) { int curr = s.top(); cout << curr << " "; s.pop(); } } // Driver code int main() { Node* root = new Node(5); root->left = new Node(9); root->right = new Node(3); root->left->left = new Node(6); root->left->right = new Node(4); root->left->left->left = new Node(8); root->left->left->right = new Node(7); reverseSpiral(root); return 0; } |
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Output:
8 7 4 6 9 3 5
Improved by :HarendraSingh22
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