Creating a tree with Left-Child Right-Sibling Representation

Left-Child Right-Sibling Representation is a different representation of an n-ary tree where instead of holding a reference to each and every child node, a node holds just two references, first a reference to it’s first child, and the other to it’s immediate next sibling. This new transformation not only removes the need of advance knowledge of the number of children a node has, but also limits the number of references to a maximum of two, thereby making it so much easier to code.

At each node, link children of same parent from left to right.
Parent should be linked with only first child.

Examples:

Left Child Right Sibling tree representation
      10
      |  
      2 -> 3 -> 4 -> 5
      |    |  
      6    7 -> 8 -> 9


Prerequisite : Left-Child Right-Sibling Representation of Tree
Below is the implementation.

C++

// CPP program to create a tree with left child
// right sibling representation.
#include<bits/stdc++.h>
using namespace std;
  
struct Node
{
    int data;
    struct Node *next;
    struct Node *child;
};
  
// Creating new Node
Node* newNode(int data)
{
    Node *newNode = new Node;
    newNode->next = newNode->child = NULL;
    newNode->data = data;
    return newNode;
}
  
// Adds a sibling to a list with starting with n
Node *addSibling(Node *n, int data)
{
    if (n == NULL)
        return NULL;
  
    while (n->next)
        n = n->next;
  
    return (n->next = newNode(data));
}
  
// Add child Node to a Node
Node *addChild(Node * n, int data)
{
    if (n == NULL)
        return NULL;
  
    // Check if child list is not empty.
    if (n->child)
        return addSibling(n->child, data);
    else
        return (n->child = newNode(data));
}
  
// Traverses tree in level order
void traverseTree(Node * root)
{
    if (root == NULL)
        return;
  
    while (root)
    {
        cout << " " << root->data;
        if (root->child)
            traverseTree(root->child);
        root = root->next;
    }
}
  
//Driver code
  
int main()
{
    /*   Let us create below tree
    *           10
    *     /   /    \   \
    *    2  3      4   5
    *              |   /  | \
    *              6   7  8  9   */
  
    // Left child right sibling
    /*  10
    *    |
    *    2 -> 3 -> 4 -> 5
    *              |    |
    *              6    7 -> 8 -> 9  */
    Node *root = newNode(10);
    Node *n1  = addChild(root, 2);
    Node *n2  = addChild(root, 3);
    Node *n3  = addChild(root, 4);
    Node *n4  = addChild(n3, 6);
    Node *n5  = addChild(root, 5);
    Node *n6  = addChild(n5, 7);
    Node *n7  = addChild(n5, 8);
    Node *n8  = addChild(n5, 9);
    traverseTree(root);
    return 0;
}

Java

// CPP program to create a tree with left child
// right sibling representation. 
  
class GFG {
      
    static class NodeTemp
    {
        int data;
        NodeTemp next, child;
        public NodeTemp(int data)
        {
            this.data = data;
            next = child = null;
        }
    }
      
    // Adds a sibling to a list with starting with n
    static public NodeTemp addSibling(NodeTemp node, int data)
    {
        if(node == null)
            return null;
        while(node.next != null)
            node = node.next;
        return(node.next = new NodeTemp(data));
    }
          
    // Add child Node to a Node
    static public NodeTemp addChild(NodeTemp node,int data)
    {
        if(node == null)
            return null;
      
        // Check if child is not empty.
        if(node.child != null)
            return(addSibling(node.child,data));
        else
            return(node.child = new NodeTemp(data));
    }
  
    // Traverses tree in level order
    static public void traverseTree(NodeTemp root)
    {
        if(root == null)
            return;
        while(root != null)
        {
            System.out.print(root.data + " ");
            if(root.child != null)
                traverseTree(root.child);
            root = root.next;
        }
    }
  
    // Driver code
    public static void main(String args[])
    {
          
        /*   Let us create below tree
        *           10
        *     /   /    \   \
        *    2  3      4   5
        *              |   /  | \
        *              6   7  8  9   */
       
        // Left child right sibling
        /*  10
        *    |
        *    2 -> 3 -> 4 -> 5
        *              |    |
        *              6    7 -> 8 -> 9  */
  
        NodeTemp root = new NodeTemp(10);
        NodeTemp n1 = addChild(root,2);
        NodeTemp n2 = addChild(root,3);
        NodeTemp n3 = addChild(root,4);
        NodeTemp n4 = addChild(n3,6);
        NodeTemp n5 = addChild(root,5);
        NodeTemp n6 = addChild(n5,7);
        NodeTemp n7 = addChild(n5,8);
        NodeTemp n8 = addChild(n5,9);
          
        traverseTree(root);
    }
}
  
// This code is contributed by M.V.S.Surya Teja.


Output:

10 2 3 4 6 5 7 8 9

This article is contributed by SAKSHI TIWARI. If you like GeeksforGeeks(We know you do!) and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : MvssTeja



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