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.

spiral_order

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Method 1 (Recursive)
This problem can bee 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 <stdio.h> 
#include <stdlib.h> 
#include <stdbool.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 traverseed 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; 
} 

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 transversed 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) 

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 transversed 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 

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 ptinting 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; 
} 

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 
        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 ptinting 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) 

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 ptinting 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 

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.

My Personal Notes

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C

Java

C#

C++

Java

C#

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