Convert left-right representation of a binary tree to down-right

Left-Right representation of a binary tree is standard representation where every node has a pointer to left child and another pointer to right child.

Down-Right representation is an alternate representation where every node has a pointer to left (or first) child and another pointer to next sibling. So siblings at every level are connected from left to right.

Given a binary tree in left-right representation as below



                               1
                    /    \
                   2      3
                   /    \
                   4      5
                     /     /   \
                    6    7      8 

Convert the structure of the tree to down-right representation like the below tree.

            1
            |
            2 – 3
                |
                4 — 5
                |   |
                6   7 – 8 

The conversion should happen in-place, i.e., left child pointer should be used as down pointer and right child pointer should be used as right sibling pointer.

We strongly recommend to minimize your browser and try this yourself.
The idea is to first convert left and right children, then convert the root. Following is C++ implementation of the idea.

C++

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/* C++ program to convert left-right to down-right
   representation of binary tree */
#include <iostream>
#include <queue>
using namespace std;
  
// A Binary Tree Node
struct node
{
    int key;
    struct node *left, *right;
};
  
// An Iterative level order traversal based function to
// convert left-right to down-right representation.
void convert(node *root)
{
    // Base Case
    if (root == NULL)  return;
  
    // Recursively convert left an right subtrees
    convert(root->left);
    convert(root->right);
  
    // If left child is NULL, make right child as left
    // as it is the first child.
    if (root->left == NULL)
       root->left = root->right;
  
    // If left child is NOT NULL, then make right child
    // as right of left child
    else
       root->left->right = root->right;
  
    // Set root's right as NULL
    root->right = NULL;
}
  
// A utility function to traverse a tree stored in
// down-right form.
void downRightTraversal(node *root)
{
    if (root != NULL)
    {
        cout << root->key << " ";
        downRightTraversal(root->right);
        downRightTraversal(root->left);
    }
}
  
// Utility function to create a new tree node
node* newNode(int key)
{
    node *temp = new node;
    temp->key = key;
    temp->left = temp->right = NULL;
    return temp;
}
  
// Driver program to test above functions
int main()
{
    // Let us create binary tree shown in above diagram
    /*
           1
         /   \
        2     3
             / \
            4   5
           /   /  \
          6   7    8
    */
    node *root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->right->left = newNode(4);
    root->right->right = newNode(5);
    root->right->left->left = newNode(6);
    root->right->right->left = newNode(7);
    root->right->right->right = newNode(8);
  
    convert(root);
  
    cout << "Traversal of the tree converted to down-right form\n";
    downRightTraversal(root);
  
    return 0;
}

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Java

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/* Java program to convert left-right to 
down-right representation of binary tree */
class GFG 
  
// A Binary Tree Node 
static class node 
    int key; 
    node left, right; 
    node(int key)
    {
        this.key = key;
        this.left = null;
        this.right = null;
    }
}
  
// An Iterative level order traversal 
// based function to convert left-right 
// to down-right representation. 
static void convert(node root) 
    // Base Case 
    if (root == null) return
  
    // Recursively convert left 
    // an right subtrees 
    convert(root.left); 
    convert(root.right); 
  
    // If left child is NULL, make right 
    // child as left as it is the first child. 
    if (root.left == null
    root.left = root.right; 
  
    // If left child is NOT NULL, then make 
    // right child as right of left child 
    else
    root.left.right = root.right; 
  
    // Set root's right as NULL 
    root.right = null
  
// A utility function to traverse a 
// tree stored in down-right form. 
static void downRightTraversal(node root) 
    if (root != null
    
        System.out.print(root.key + " "); 
        downRightTraversal(root.right); 
        downRightTraversal(root.left); 
    
  
// Utility function to create
// a new tree node 
static node newNode(int key) 
    node temp = new node(0); 
    temp.key = key; 
    temp.left = null;
    temp.right = null
    return temp; 
  
// Driver Code
public static void main(String[] args) 
    // Let us create binary tree
    // shown in above diagram 
    /* 
        
        / \ 
        2 3 
            / \ 
            4 5 
        / / \ 
        6 7 8 
    */
    node root = new node(1);
    root.left = newNode(2); 
    root.right = newNode(3); 
    root.right.left = newNode(4); 
    root.right.right = newNode(5); 
    root.right.left.left = newNode(6); 
    root.right.right.left = newNode(7); 
    root.right.right.right = newNode(8); 
  
    convert(root); 
  
    System.out.println("Traversal of the tree "
                 "converted to down-right form"); 
    downRightTraversal(root); 
}
  
// This code is contributed
// by Prerna Saini

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Python3

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# Python3 program to convert left-right to
# down-right representation of binary tree
  
# Helper function that allocates a new 
# node with the given data and None 
# left and right poers.                                     
class newNode: 
  
    # Construct to create a new node 
    def __init__(self, key): 
        self.key = key
        self.left = None
        self.right = None
  
# An Iterative level order traversal based 
# function to convert left-right to down-right
# representation.
def convert(root):
  
    # Base Case
    if (root == None):
        return
  
    # Recursively convert left an 
    # right subtrees
    convert(root.left)
    convert(root.right)
  
    # If left child is None, make right 
    # child as left as it is the first child.
    if (root.left == None):
        root.left = root.right
  
    # If left child is NOT None, then make
    # right child as right of left child
    else:
        root.left.right = root.right
  
    # Set root's right as None
    root.right = None
  
# A utility function to traverse a 
# tree stored in down-right form.
def downRightTraversal(root):
  
    if (root != None):
      
        print( root.key, end = " ")
        downRightTraversal(root.right)
        downRightTraversal(root.left)
  
# Driver Code 
if __name__ == '__main__':
      
    # Let us create binary tree shown
    # in above diagram
    """
        1
        / \
        2     3
            / \
            4 5
        / / \
        6 7 8
    """
    root = newNode(1)
    root.left = newNode(2)
    root.right = newNode(3)
    root.right.left = newNode(4)
    root.right.right = newNode(5)
    root.right.left.left = newNode(6)
    root.right.right.left = newNode(7)
    root.right.right.right = newNode(8)
  
    convert(root)
      
    print("Traversal of the tree converted"
                       "to down-right form")
    downRightTraversal(root)
  
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

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C#

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// C# program to convert left-right to 
// down-right representation of binary tree 
using System;
  
class GFG
{
  
// A Binary Tree Node 
public class node
{
    public int key;
    public node left, right;
    public node(int key)
    {
        this.key = key;
        this.left = null;
        this.right = null;
    }
}
  
// An Iterative level order traversal 
// based function to convert left-right 
// to down-right representation. 
public static void convert(node root)
{
    // Base Case 
    if (root == null)
    {
        return;
    }
  
    // Recursively convert left 
    // an right subtrees 
    convert(root.left);
    convert(root.right);
  
    // If left child is NULL, make right 
    // child as left as it is the first child. 
    if (root.left == null)
    {
        root.left = root.right;
    }
  
    // If left child is NOT NULL, then make 
    // right child as right of left child 
    else
    {
        root.left.right = root.right;
    }
  
    // Set root's right as NULL 
    root.right = null;
}
  
// A utility function to traverse a 
// tree stored in down-right form. 
public static void downRightTraversal(node root)
{
    if (root != null)
    {
        Console.Write(root.key + " ");
        downRightTraversal(root.right);
        downRightTraversal(root.left);
    }
}
  
// Utility function to create 
// a new tree node 
public static node newNode(int key)
{
    node temp = new node(0);
    temp.key = key;
    temp.left = null;
    temp.right = null;
    return temp;
}
  
// Driver Code 
public static void Main(string[] args)
{
    // Let us create binary tree 
    // shown in above diagram 
    /* 
        
        / \ 
        2 3 
            / \ 
            4 5 
        / / \ 
        6 7 8 
    */
    node root = new node(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.right.left = newNode(4);
    root.right.right = newNode(5);
    root.right.left.left = newNode(6);
    root.right.right.left = newNode(7);
    root.right.right.right = newNode(8);
  
    convert(root);
  
    Console.WriteLine("Traversal of the tree "
                      "converted to down-right form");
    downRightTraversal(root);
}
}
  
// This code is contributed 
// by Shrikant13

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Output:

Traversal of the tree converted to down-right form
1 2 3 4 5 7 8 6

Time complexity of the above program is O(n).

This article is contributed by Abhishek. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



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