Level order traversal in spiral form
Write a function to print spiral order traversal of a tree. For below tree, function should print 1, 2, 3, 4, 5, 6, 7.
Method 1 (Recursive)
This problem can be seen as an extension of the level order traversal post.
To print the nodes in spiral order, nodes at different levels should be printed in alternating order. An additional Boolean variable ltr is used to change printing order of levels. If ltr is 1 then printGivenLevel() prints nodes from left to right else from right to left. Value of ltr is flipped in each iteration to change the order.
Function to print level order traversal of tree
printSpiral(tree)
bool ltr = 0;
for d = 1 to height(tree)
printGivenLevel(tree, d, ltr);
ltr ~= ltr /*flip ltr*/
Function to print all nodes at a given level
printGivenLevel(tree, level, ltr)
if tree is NULL then return;
if level is 1, then
print(tree->data);
else if level greater than 1, then
if(ltr)
printGivenLevel(tree->left, level-1, ltr);
printGivenLevel(tree->right, level-1, ltr);
else
printGivenLevel(tree->right, level-1, ltr);
printGivenLevel(tree->left, level-1, ltr);
Following is the implementation of above algorithm.
C
// C program for recursive level order traversal in spiral form #include <stdbool.h> #include <stdio.h> #include <stdlib.h> /* A binary tree node has data, pointer to left child and a pointer to right child */struct node { int data; struct node* left; struct node* right; }; /* Function protoypes */void printGivenLevel(struct node* root, int level, int ltr); int height(struct node* node); struct node* newNode(int data); /* Function to print spiral traversal of a tree*/void printSpiral(struct node* root) { int h = height(root); int i; /*ltr -> Left to Right. If this variable is set, then the given level is traversed from left to right. */ bool ltr = false; for (i = 1; i <= h; i++) { printGivenLevel(root, i, ltr); /*Revert ltr to traverse next level in opposite order*/ ltr = !ltr; } } /* Print nodes at a given level */void printGivenLevel(struct node* root, int level, int ltr) { if (root == NULL) return; if (level == 1) printf("%d ", root->data); else if (level > 1) { if (ltr) { printGivenLevel(root->left, level - 1, ltr); printGivenLevel(root->right, level - 1, ltr); } else { printGivenLevel(root->right, level - 1, ltr); printGivenLevel(root->left, level - 1, ltr); } } } /* Compute the "height" of a tree -- the number of nodes along the longest path from the root node down to the farthest leaf node.*/int height(struct node* node) { if (node == NULL) return 0; else { /* compute the height of each subtree */ int lheight = height(node->left); int rheight = height(node->right); /* use the larger one */ if (lheight > rheight) return (lheight + 1); else return (rheight + 1); } } /* Helper function that allocates a new node with the given data and NULL left and right pointers. */struct node* newNode(int data) { struct node* node = (struct node*) malloc(sizeof(struct node)); node->data = data; node->left = NULL; node->right = NULL; return (node); } /* Driver program to test above functions*/int main() { struct node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(7); root->left->right = newNode(6); root->right->left = newNode(5); root->right->right = newNode(4); printf("Spiral Order traversal of binary tree is \n"); printSpiral(root); return 0; } |
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Java
// Java program for recursive level order traversal in spiral form /* A binary tree node has data, pointer to left child and a pointer to right child */class Node { int data; Node left, right; public Node(int d) { data = d; left = right = null; } } class BinaryTree { Node root; // Function to print the spiral traversal of tree void printSpiral(Node node) { int h = height(node); int i; /* ltr -> left to right. If this variable is set then the given label is traversed from left to right */ boolean ltr = false; for (i = 1; i <= h; i++) { printGivenLevel(node, i, ltr); /*Revert ltr to traverse next level in opposite order*/ ltr = !ltr; } } /* Compute the "height" of a tree -- the number of nodes along the longest path from the root node down to the farthest leaf node.*/ int height(Node node) { if (node == null) return 0; else { /* compute the height of each subtree */ int lheight = height(node.left); int rheight = height(node.right); /* use the larger one */ if (lheight > rheight) return (lheight + 1); else return (rheight + 1); } } /* Print nodes at a given level */ void printGivenLevel(Node node, int level, boolean ltr) { if (node == null) return; if (level == 1) System.out.print(node.data + " "); else if (level > 1) { if (ltr != false) { printGivenLevel(node.left, level - 1, ltr); printGivenLevel(node.right, level - 1, ltr); } else { printGivenLevel(node.right, level - 1, ltr); printGivenLevel(node.left, level - 1, ltr); } } } /* Driver program to test the above functions */ public static void main(String[] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(7); tree.root.left.right = new Node(6); tree.root.right.left = new Node(5); tree.root.right.right = new Node(4); System.out.println("Spiral order traversal of Binary Tree is "); tree.printSpiral(tree.root); } } // This code has been contributed by Mayank Jaiswal(mayank_24) |
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Python3
# Python3 program for recursive level order # traversal in spiral form class newNode: # Construct to create a newNode def __init__(self, key): self.data = key self.left = None self.right = None """ Function to print spiral traversal of a tree"""def printSpiral(root): h = height(root) """ltr Left to Right. If this variable is set, then the given level is traversed from left to right. """ ltr = False for i in range(1, h + 1): printGivenLevel(root, i, ltr) """Revert ltr to traverse next level in opposite order""" ltr = not ltr """ Print nodes at a given level """def printGivenLevel(root, level, ltr): if(root == None): return if(level == 1): print(root.data, end = " ") elif (level > 1): if(ltr): printGivenLevel(root.left, level - 1, ltr) printGivenLevel(root.right, level - 1, ltr) else: printGivenLevel(root.right, level - 1, ltr) printGivenLevel(root.left, level - 1, ltr) """ Compute the "height" of a tree -- the number of nodes along the longest path from the root node down to the farthest leaf node."""def height(node): if (node == None): return 0 else: """ compute the height of each subtree """ lheight = height(node.left) rheight = height(node.right) """ use the larger one """ if (lheight > rheight): return(lheight + 1) else: return(rheight + 1) # Driver Code if __name__ == '__main__': root = newNode(1) root.left = newNode(2) root.right = newNode(3) root.left.left = newNode(7) root.left.right = newNode(6) root.right.left = newNode(5) root.right.right = newNode(4) print("Spiral Order traversal of binary tree is") printSpiral(root) # This code is contributed # by SHUBHAMSINGH10 |
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C#
// C# program for recursive level // order traversal in spiral form using System; /* 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; public Node(int d) { data = d; left = right = null; } } class GFG { public Node root; // Function to print the spiral // traversal of tree public virtual void printSpiral(Node node) { int h = height(node); int i; /* ltr -> left to right. If this variable is set then the given label is traversed from left to right */ bool ltr = false; for (i = 1; i <= h; i++) { printGivenLevel(node, i, ltr); /*Revert ltr to traverse next level in opposite order*/ ltr = !ltr; } } /* Compute the "height" of a tree -- the number of nodes along the longest path from the root node down to the farthest leaf node.*/ public virtual int height(Node node) { if (node == null) { return 0; } else { /* compute the height of each subtree */ int lheight = height(node.left); int rheight = height(node.right); /* use the larger one */ if (lheight > rheight) { return (lheight + 1); } else { return (rheight + 1); } } } /* Print nodes at a given level */ public virtual void printGivenLevel(Node node, int level, bool ltr) { if (node == null) { return; } if (level == 1) { Console.Write(node.data + " "); } else if (level > 1) { if (ltr != false) { printGivenLevel(node.left, level - 1, ltr); printGivenLevel(node.right, level - 1, ltr); } else { printGivenLevel(node.right, level - 1, ltr); printGivenLevel(node.left, level - 1, ltr); } } } // Driver Code public static void Main(string[] args) { GFG tree = new GFG(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(7); tree.root.left.right = new Node(6); tree.root.right.left = new Node(5); tree.root.right.right = new Node(4); Console.WriteLine("Spiral order traversal " + "of Binary Tree is "); tree.printSpiral(tree.root); } } // This code is contributed by Shrikant13 |
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Output:
Spiral Order traversal of binary tree is 1 2 3 4 5 6 7
Time Complexity: Worst case time complexity of the above method is O(n^2). Worst case occurs in case of skewed trees.
Method 2 (Iterative)
We can print spiral order traversal in O(n) time and O(n) extra space. The idea is to use two stacks. We can use one stack for printing from left to right and other stack for printing from right to left. In every iteration, we have nodes of one level in one of the stacks. We print the nodes, and push nodes of next level in other stack.
C++
// C++ implementation of a O(n) time method for spiral order traversal #include <iostream> #include <stack> using namespace std; // Binary Tree node struct node { int data; struct node *left, *right; }; void printSpiral(struct node* root) { if (root == NULL) return; // NULL check // Create two stacks to store alternate levels stack<struct node*> s1; // For levels to be printed from right to left stack<struct node*> s2; // For levels to be printed from left to right // Push first level to first stack 's1' s1.push(root); // Keep printing while any of the stacks has some nodes while (!s1.empty() || !s2.empty()) { // Print nodes of current level from s1 and push nodes of // next level to s2 while (!s1.empty()) { struct node* temp = s1.top(); s1.pop(); cout << temp->data << " "; // Note that is right is pushed before left if (temp->right) s2.push(temp->right); if (temp->left) s2.push(temp->left); } // Print nodes of current level from s2 and push nodes of // next level to s1 while (!s2.empty()) { struct node* temp = s2.top(); s2.pop(); cout << temp->data << " "; // Note that is left is pushed before right if (temp->left) s1.push(temp->left); if (temp->right) s1.push(temp->right); } } } // A utility function to create a new node struct node* newNode(int data) { struct node* node = new struct node; node->data = data; node->left = NULL; node->right = NULL; return (node); } int main() { struct node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(7); root->left->right = newNode(6); root->right->left = newNode(5); root->right->right = newNode(4); cout << "Spiral Order traversal of binary tree is \n"; printSpiral(root); return 0; } |
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Java
// Java implementation of an O(n) approach of level order // traversal in spiral form import java.util.*; // A Binary Tree node class Node { int data; Node left, right; public Node(int item) { data = item; left = right = null; } } class BinaryTree { static Node root; void printSpiral(Node node) { if (node == null) return; // NULL check // Create two stacks to store alternate levels // For levels to be printed from right to left Stack<Node> s1 = new Stack<Node>(); // For levels to be printed from left to right Stack<Node> s2 = new Stack<Node>(); // Push first level to first stack 's1' s1.push(node); // Keep printing while any of the stacks has some nodes while (!s1.empty() || !s2.empty()) { // Print nodes of current level from s1 and push nodes of // next level to s2 while (!s1.empty()) { Node temp = s1.peek(); s1.pop(); System.out.print(temp.data + " "); // Note that is right is pushed before left if (temp.right != null) s2.push(temp.right); if (temp.left != null) s2.push(temp.left); } // Print nodes of current level from s2 and push nodes of // next level to s1 while (!s2.empty()) { Node temp = s2.peek(); s2.pop(); System.out.print(temp.data + " "); // Note that is left is pushed before right if (temp.left != null) s1.push(temp.left); if (temp.right != null) s1.push(temp.right); } } } public static void main(String[] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(7); tree.root.left.right = new Node(6); tree.root.right.left = new Node(5); tree.root.right.right = new Node(4); System.out.println("Spiral Order traversal of Binary Tree is "); tree.printSpiral(root); } } // This code has been contributed by Mayank Jaiswal(mayank_24) |
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Python3
# Python3 implementation of a O(n) time # method for spiral order traversal # A class to create a new node class newNode: def __init__(self, data): self.data = data self.left = None self.right = None def printSpiral(root): if (root == None): return # None check # Create two stacks to store # alternate levels s1 = [] # For levels to be printed # from right to left s2 = [] # For levels to be printed # from left to right # append first level to first stack 's1' s1.append(root) # Keep printing while any of the # stacks has some nodes while not len(s1) == 0 or not len(s2) == 0: # Print nodes of current level from s1 # and append nodes of next level to s2 while not len(s1) == 0: temp = s1[-1] s1.pop() print(temp.data, end = " ") # Note that is right is appended # before left if (temp.right): s2.append(temp.right) if (temp.left): s2.append(temp.left) # Print nodes of current level from s2 # and append nodes of next level to s1 while (not len(s2) == 0): temp = s2[-1] s2.pop() print(temp.data, end = " ") # Note that is left is appended # before right if (temp.left): s1.append(temp.left) if (temp.right): s1.append(temp.right) # Driver Code if __name__ == '__main__': root = newNode(1) root.left = newNode(2) root.right = newNode(3) root.left.left = newNode(7) root.left.right = newNode(6) root.right.left = newNode(5) root.right.right = newNode(4) print("Spiral Order traversal of", "binary tree is ") printSpiral(root) # This code is contributed by PranchalK |
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C#
// C# implementation of an O(n) approach of // level order traversal in spiral form using System; using System.Collections.Generic; // A Binary Tree node public class Node { public int data; public Node left, right; public Node(int item) { data = item; left = right = null; } } public class BinaryTree { public static Node root; public virtual void printSpiral(Node node) { if (node == null) { return; // NULL check } // Create two stacks to store alternate levels Stack<Node> s1 = new Stack<Node>(); // For levels to be printed // from right to left Stack<Node> s2 = new Stack<Node>(); // For levels to be printed // from left to right // Push first level to first stack 's1' s1.Push(node); // Keep printing while any of the // stacks has some nodes while (s1.Count > 0 || s2.Count > 0) { // Print nodes of current level from // s1 and push nodes of next level to s2 while (s1.Count > 0) { Node temp = s1.Peek(); s1.Pop(); Console.Write(temp.data + " "); // Note that is right is pushed before left if (temp.right != null) { s2.Push(temp.right); } if (temp.left != null) { s2.Push(temp.left); } } // Print nodes of current level from s2 // and push nodes of next level to s1 while (s2.Count > 0) { Node temp = s2.Peek(); s2.Pop(); Console.Write(temp.data + " "); // Note that is left is pushed before right if (temp.left != null) { s1.Push(temp.left); } if (temp.right != null) { s1.Push(temp.right); } } } } // Driver Code public static void Main(string[] args) { BinaryTree tree = new BinaryTree(); BinaryTree.root = new Node(1); BinaryTree.root.left = new Node(2); BinaryTree.root.right = new Node(3); BinaryTree.root.left.left = new Node(7); BinaryTree.root.left.right = new Node(6); BinaryTree.root.right.left = new Node(5); BinaryTree.root.right.right = new Node(4); Console.WriteLine("Spiral Order traversal of Binary Tree is "); tree.printSpiral(root); } } // This code is contributed by Shrikant13 |
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Output:
Spiral Order traversal of binary tree is 1 2 3 4 5 6 7
Please write comments if you find any bug in the above program/algorithm; or if you want to share more information about spiral traversal.
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