Double Order Traversal of a Binary Tree
Last Updated :
11 Oct, 2021
Given a Binary Tree consisting of N nodes, the task is to print its Double Order Traversal.
Double Order Traversal is a tree traversal technique in which every node is traversed twice in the following order:
- Visit the Node.
- Traverse the Left Subtree.
- Visit the Node.
- Traverse the Right Subtree.
Examples:
Input:
1
/ \
7 3
/ \ /
4 5 6
Output: 1 7 4 4 7 5 5 1 3 6 6 3
Input:
1
/ \
7 3
/ \ \
4 5 6
Output: 1 7 4 4 7 5 5 1 3 3 6 6
Approach:
The idea is to perform Inorder Traversal recursively on the given Binary Tree and print the node value on visiting a vertex and after the recursive call to the left subtree during the traversal.
Follow the steps below to solve the problem:
- Start Inorder traversal from the root.
- If the current node does not exist, simply return from it.
- Otherwise:
- Print the value of the current node.
- Recursively traverse the left subtree.
- Again, print the current node.
- Recursively traverse the right subtree.
- Repeat the above steps until all nodes in the tree are visited.
Below is the implementation of the above approach:
C++
#include <iostream>
using namespace std;
struct node {
char data;
struct node *left, *right;
};
struct node* newNode( char ch)
{
struct node* Node = new node();
Node->data = ch;
Node->left = NULL;
Node->right = NULL;
return Node;
}
void doubleOrderTraversal( struct node* root)
{
if (!root)
return ;
cout << root->data << " " ;
doubleOrderTraversal(root->left);
cout << root->data << " " ;
doubleOrderTraversal(root->right);
}
int main()
{
struct node* root = newNode( '1' );
root->left = newNode( '7' );
root->right = newNode( '3' );
root->left->left = newNode( '4' );
root->left->right = newNode( '5' );
root->right->right = newNode( '6' );
doubleOrderTraversal(root);
return 0;
}
|
Java
class GFG{
static class node
{
char data;
node left, right;
};
static node newNode( char ch)
{
node n = new node();
n.data = ch;
n.left = null ;
n.right = null ;
return n;
}
static void doubleOrderTraversal(node root)
{
if (root == null )
return ;
System.out.print(root.data + " " );
doubleOrderTraversal(root.left);
System.out.print(root.data + " " );
doubleOrderTraversal(root.right);
}
public static void main(String[] args)
{
node root = newNode( '1' );
root.left = newNode( '7' );
root.right = newNode( '3' );
root.left.left = newNode( '4' );
root.left.right = newNode( '5' );
root.right.right = newNode( '6' );
doubleOrderTraversal(root);
}
}
|
Python3
class Node:
def __init__( self , ch):
self .data = ch
self .left = None
self .right = None
def doubleOrderTraveersal(root):
if not root:
return
print (root.data, end = " " )
doubleOrderTraveersal(root.left)
print (root.data, end = " " )
doubleOrderTraveersal(root.right)
if __name__ = = '__main__' :
root = Node( 1 )
root.left = Node( 7 )
root.right = Node( 3 )
root.left.left = Node( 4 )
root.left.right = Node( 5 )
root.right.right = Node( 6 )
doubleOrderTraveersal(root)
|
C#
using System;
class GFG{
class node
{
public char data;
public node left, right;
};
static node newNode( char ch)
{
node n = new node();
n.data = ch;
n.left = null ;
n.right = null ;
return n;
}
static void doubleOrderTraversal(node root)
{
if (root == null )
return ;
Console.Write(root.data + " " );
doubleOrderTraversal(root.left);
Console.Write(root.data + " " );
doubleOrderTraversal(root.right);
}
public static void Main(String[] args)
{
node root = newNode( '1' );
root.left = newNode( '7' );
root.right = newNode( '3' );
root.left.left = newNode( '4' );
root.left.right = newNode( '5' );
root.right.right = newNode( '6' );
doubleOrderTraversal(root);
}
}
|
Javascript
<script>
class node
{
constructor()
{
this .data = 0;
this .left = null ;
this .right = null ;
}
};
function newNode(ch)
{
var n = new node();
n.data = ch;
n.left = null ;
n.right = null ;
return n;
}
function doubleOrderTraversal(root)
{
if (root == null )
return ;
document.write(root.data + " " );
doubleOrderTraversal(root.left);
document.write(root.data + " " );
doubleOrderTraversal(root.right);
}
var root = newNode( '1' );
root.left = newNode( '7' );
root.right = newNode( '3' );
root.left.left = newNode( '4' );
root.left.right = newNode( '5' );
root.right.right = newNode( '6' );
doubleOrderTraversal(root);
</script>
|
Output:
1 7 4 4 7 5 5 1 3 3 6 6
Time Complexity: O(N)
Auxiliary Space: O(1)
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