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Code Recursion

# Double order traversal of a binary tree

The problem involves performing a double-order traversal of a binary tree. In this traversal, each node is visited twice: first in a preorder traversal, then in an inorder traversal. The goal is to print the nodes during each traversal, resulting in a sequence of nodes that appears twice in the specified order.

## Problem Statement

Given a binary tree, the task is to perform a double-order traversal. During the traversal, each node is visited twice: first in a preorder fashion, and then in an inorder fashion. The program should print the nodes during each traversal.

## Example

Consider the binary tree given in the code:

``````         4
/ \
/   \
-4     7
/ \     \
2   3     1
/ \   /
6   8 5
/
9``````

The double-order traversal output for this tree will be:

``4 -4 2 2 -4 3 6 9 9 6 3 8 8 4 7 7 1 5 5 1``

## Idea to Solve

The problem can be solved using a recursive approach. During the traversal, at each node, we perform the following steps:

1. Print the node value during the preorder traversal.
2. Recursively traverse the left subtree.
3. Print the node value during the inorder traversal.
4. Recursively traverse the right subtree.

By following these steps, we ensure that each node is visited twice in the order of preorder and inorder traversal.

## Pseudocode

``````function doubleOrder(node):
if node is not null:
print node.data during preorder
doubleOrder(node.left)
print node.data during inorder
doubleOrder(node.right)

main:
create binary tree
construct tree as shown
call doubleOrder with root node``````

## Algorithm Explanation

1. The `doubleOrder` function recursively traverses the binary tree and performs the specified actions during each traversal type (preorder and inorder).
2. In the `main` function, a binary tree is constructed, and the `doubleOrder` function is called with the root node.

## Time Complexity

The time complexity of the solution is proportional to the number of nodes in the binary tree since each node is visited once during the traversal. In the worst case, the time complexity is O(n), where n is the number of nodes in the binary tree.

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