# Construct Complete Binary Tree from its Linked List Representation

Given Linked List Representation of Complete Binary Tree, construct the Binary tree. A complete binary tree can be represented in an array in the following approach.

If root node is stored at index i, its left, and right children are stored at indices 2*i+1, 2*i+2 respectively.
Suppose tree is represented by a linked list in same way, how do we convert this into normal linked representation of binary tree where every node has data, left and right pointers? In the linked list representation, we cannot directly access the children of the current node unless we traverse the list.

We are mainly given level order traversal in sequential access form. We know head of linked list is always is root of the tree. We take the first node as root and we also know that the next two nodes are left and right children of root. So we know partial Binary Tree. The idea is to do Level order traversal of the partially built Binary Tree using queue and traverse the linked list at the same time. At every step, we take the parent node from queue, make next two nodes of linked list as children of the parent node, and enqueue the next two nodes to queue.

1. Create an empty queue.
2. Make the first node of the list as root, and enqueue it to the queue.
3. Until we reach the end of the list, do the following.
………a. Dequeue one node from the queue. This is the current parent.
………b. Traverse two nodes in the list, add them as children of the current parent.
………c. Enqueue the two nodes into the queue.

Below is the code which implements the same in C++.

## C++

// C++ program to create a Complete Binary tree from its Linked List
// Representation
#include <iostream>
#include <string>
#include <queue>
using namespace std;

struct ListNode
{
int data;
ListNode* next;
};

// Binary tree node structure
struct BinaryTreeNode
{
int data;
BinaryTreeNode *left, *right;
};

// Function to insert a node at the beginning of the Linked List
void push(struct ListNode** head_ref, int new_data)
{
// allocate node and assign data
struct ListNode* new_node = new ListNode;
new_node->data = new_data;

// link the old list off the new node

// move the head to point to the new node
}

// method to create a new binary tree node from the given data
BinaryTreeNode* newBinaryTreeNode(int data)
{
BinaryTreeNode *temp = new BinaryTreeNode;
temp->data = data;
temp->left = temp->right = NULL;
return temp;
}

// converts a given linked list representing a complete binary tree into the
// linked representation of binary tree.
{
// queue to store the parent nodes
queue<BinaryTreeNode *> q;

// Base Case
{
root = NULL; // Note that root is passed by reference
return;
}

// 1.) The first node is always the root node, and add it to the queue
q.push(root);

// advance the pointer to the next node

// until the end of linked list is reached, do the following steps
{
// 2.a) take the parent node from the q and remove it from q
BinaryTreeNode* parent = q.front();
q.pop();

// 2.c) take next two nodes from the linked list. We will add
// them as children of the current parent node in step 2.b. Push them
// into the queue so that they will be parents to the future nodes
BinaryTreeNode *leftChild = NULL, *rightChild = NULL;
q.push(leftChild);
{
q.push(rightChild);
}

// 2.b) assign the left and right children of parent
parent->left = leftChild;
parent->right = rightChild;
}
}

// Utility function to traverse the binary tree after conversion
void inorderTraversal(BinaryTreeNode* root)
{
if (root)
{
inorderTraversal( root->left );
cout << root->data << " ";
inorderTraversal( root->right );
}
}

// Driver program to test above functions
int main()
{
// create a linked list shown in above diagram

BinaryTreeNode *root;

cout << "Inorder Traversal of the constructed Binary Tree is: \n";
inorderTraversal(root);
return 0;
}

## Java

// Java program to create complete Binary Tree from its Linked List
// representation

// importing necessary classes
import java.util.*;

class ListNode
{
int data;
ListNode next;
ListNode(int d)
{
data = d;
next = null;
}
}

// A binary tree node
class BinaryTreeNode
{
int data;
BinaryTreeNode left, right = null;
BinaryTreeNode(int data)
{
this.data = data;
left = right = null;
}
}

class BinaryTree
{
BinaryTreeNode root;

// Function to insert a node at the beginning of
void push(int new_data)
{
// allocate node and assign data
ListNode new_node = new ListNode(new_data);

// link the old list off the new node

// move the head to point to the new node
}

// converts a given linked list representing a
// complete binary tree into the linked
// representation of binary tree.
BinaryTreeNode convertList2Binary(BinaryTreeNode node)
{
// queue to store the parent nodes
Queue<BinaryTreeNode> q =

// Base Case
{
node = null;
return null;
}

// 1.) The first node is always the root node, and
//     add it to the queue

// advance the pointer to the next node

// until the end of linked list is reached, do the
// following steps
{
// 2.a) take the parent node from the q and
//      remove it from q
BinaryTreeNode parent = q.peek();
BinaryTreeNode pp = q.poll();

// 2.c) take next two nodes from the linked list.
// We will add them as children of the current
// parent node in step 2.b. Push them into the
// queue so that they will be parents to the
// future nodes
BinaryTreeNode leftChild = null, rightChild = null;
{
}

// 2.b) assign the left and right children of
//      parent
parent.left = leftChild;
parent.right = rightChild;
}

return node;
}

// Utility function to traverse the binary tree
// after conversion
void inorderTraversal(BinaryTreeNode node)
{
if (node != null)
{
inorderTraversal(node.left);
System.out.print(node.data + " ");
inorderTraversal(node.right);
}
}

// Driver program to test above functions
public static void main(String[] args)
{
BinaryTree tree = new BinaryTree();
tree.push(36); /* Last node of Linked List */
tree.push(30);
tree.push(25);
tree.push(15);
tree.push(12);
tree.push(10); /* First node of Linked List */
BinaryTreeNode node = tree.convertList2Binary(tree.root);

System.out.println("Inorder Traversal of the"+
" constructed Binary Tree is:");
tree.inorderTraversal(node);
}
}
// This code has been contributed by Mayank Jaiswal

## Python

# Python program to create a Complete Binary Tree from

class ListNode:

# Constructor to create a new node
def __init__(self, data):
self.data = data
self.next = None

# Binary Tree Node structure
class BinaryTreeNode:

# Constructor to create a new node
def __init__(self, data):
self.data = data
self.left = None
self.right = None

# Class to convert the linked list to Binary Tree
class Conversion:

# and root for the Binary Tree
def __init__(self, data = None):
self.root = None

def push(self, new_data):

# Creating a new linked list node and storing data
new_node = ListNode(new_data)

# Make next of new node as head

# Move the head to point to new node

def convertList2Binary(self):

# Queue to store the parent nodes
q = []

# Base Case
self.root = None
return

# 1.) The first node is always the root node,
# and add it to the queue
q.append(self.root)

# Advance the pointer to the next node

# Until th end of linked list is reached, do:

# 2.a) Take the parent node from the q and
# and remove it from q
parent = q.pop(0) # Front of queue

# 2.c) Take next two nodes from the linked list.
# We will add them as children of the current
# parent node in step 2.b.
# Push them into the queue so that they will be
# parent to the future node
leftChild= None
rightChild = None

q.append(leftChild)
q.append(rightChild)

#2.b) Assign the left and right children of parent
parent.left = leftChild
parent.right = rightChild

def inorderTraversal(self, root):
if(root):
self.inorderTraversal(root.left)
print root.data,
self.inorderTraversal(root.right)

# Driver Program to test above function

# Object of conversion class
conv = Conversion()
conv.push(36)
conv.push(30)
conv.push(25)
conv.push(15)
conv.push(12)
conv.push(10)

conv.convertList2Binary()

print "Inorder Traversal of the contructed Binary Tree is:"
conv.inorderTraversal(conv.root)

# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

Output:

Inorder Traversal of the constructed Binary Tree is:
25 12 30 10 36 15

Time Complexity: Time complexity of the above solution is O(n) where n is the number of nodes.

# Company Wise Coding Practice    Topic Wise Coding Practice

• ms

I guess this can be an alternate soln(Pls share ur views):
Take two pointers: root & child.
root points two the first node of the list and child to the second node.
make child and child->next the left and right children of root
progress root by one and child by two nodes..

Base cases can be handled accordingly..

• ms

Complexity O(n)

• ANA

nice trick π

• samthebest

{
return NULL;

queue q;

btnode *current=root;
q.push(root);
while(ptr!=NULL)
{ btnode *node=new_node(ptr->data);
if(!current->lchild)
{
current->lchild=node;
}
else if(!current->rchild)
{ current->rchild=node;
}

q.push(node);
if(current->lchild && current->rchild)
{
q.pop();
current=q.front();
}
ptr=ptr->next;
}

return root;
}

• prashant jha

0(n^2) can be done by using recursion passing level and node values to the recursive function which recur at 2i+1 and 2i+2 and if its value is >currrent index then return……..
0(n) by using queue
create root and enqueue it
while q.front!=-1
deque from front
check if its both child exist or not by checking 2i+1 and 2i+2 both <n
if they exist form nodes and enqueue it else enque only lchild if 2i+2 <n

• monu

hi

• sunil

Another way would be to count the number of nodes > send half the nodes to left and the other half to right(if odd number send the extra one to left)
recursively build the left subtree (termination condition would be when there are no more nodes) and right subtree (while building the tree keep sending the next node up the tree for building the right subtree).

• Dave

Call me crazy but isn’t a binary tree supposed to be sorted?

• SYED IMRAN

#include
#include
#include
struct tnode
{
int data;
struct tnode *left,*right;
};
struct node
{
int data;
struct node *next;
};
typedef struct tnode tnode;
typedef struct node node;
void push(node **h,int t)
{
node * nw=(node *)malloc(sizeof(node));
nw->data=t;
nw->next=NULL;
if(*h==NULL)
{
*h=nw;
}
else
{
nw->next=*h;
*h=nw;
}
}
tnode * newtnode(int n)
{
tnode *nw=(tnode *)malloc(sizeof(tnode));
nw->data=n;
nw->left=nw->right=NULL;
return nw;
}
int getlistnode(node *h,int i)
{
node * s;
s=h;
while(i>0)
{
s=s->next;
i–;
}
return s->data;
}
tnode * createtree(int i,int size)
{
if(i>=size) return NULL;
tnode * root=(tnode *)malloc(sizeof(tnode));
root->left=createtree(2*i+1,size);
root->right=createtree(2*i+2,size);
return root;
}
void inorderTraversal(tnode * root)
{
if (root)
{
inorderTraversal( root->left );
printf(“%d “,root->data);
inorderTraversal( root->right );
}
}
int main()
{
tnode * root=NULL;
root=createtree(0,6);
printf(“inorder is:\n”);
inorderTraversal(root);

return 0;
}

• sandeep kothapally

i have a code table like for example
2 assigned as “10”
3 assigned as “010”
4 assigned as “0110”
5 assigned as “0111”

now i wanted to write it as a binary tree so that if i input
10010 i could traverse the tree and decode it as 2 3
my query is to how to design such tree so that i can use it at the decoder

• nilesh

consider this:
{
List *ptr;
while(i++next;
}
if(!ptr)
return;
struct node *root=newnode(ptr->data);
return root;
}

• AMIT

your solution uses o(n) time and o(n) space
a similar but much easier solution, I think–
1. copy all the elements of list into an array
2. build the binary tree
this will also take o(n) time and o(n)space

• EOF

I was going to comment the same, but saw yours … π

• abhishek08aug

Intelligent π

• Vimal

There is one bug in the above code.
Queue implementation is not correct.
After one iteration of while loop above , queue would be
{15,12} but it should be { 12,15}
Correct me if i am wrong.

• neon007

#include
#include

typedef struct list {
int data;
struct list * next;
} list;

typedef struct tree {
int data;
struct tree * left;
struct tree * right;
struct tree * next;
} tree;

typedef struct next {
struct next * next;
} next;

typedef struct queue {
tree * front;
tree * rear;
} queue;

list * temp = (list *) malloc(sizeof(list));
temp->data = d;
temp->next = NULL;
} else {
}
}

printf(“inside removeItem\n”);
int item;
return -1;
return item;
}

void enque(queue ** Q, tree * t_node) {
printf(“inside enque\n”);
if (((*Q)->front == NULL) && ((*Q)->rear == NULL)) {
(*Q)->front = (tree *) malloc(sizeof(tree));
(*Q)->rear = (tree *) malloc(sizeof(tree));
(*Q)->front = t_node;
(*Q)->rear = t_node;
} else {
(*Q)->rear->next = t_node;
(*Q)->rear = t_node;
}
}

void dequeue(queue ** Q, tree ** item) {
printf(“inside deque\n”);
if ((*Q)->front == NULL) {
printf(“Queue underflow\n”);
return;
}
*item = (*Q)->front;
(*Q)->front = (*Q)->front->next;
if ((*Q)->front == NULL)
(*Q)->rear = NULL;
}

void inorderTraversal(tree ** root) {
if((*root) == NULL)
return;
inorderTraversal(&((*root)->left));
printf(“data = %d\n”, (*root)->data);
inorderTraversal(&((*root)->right));
}

void createTree(list ** head, tree * root) {
int item, left, right;
tree * temp_l, *temp_r, *node, *ret;

queue * Q = (queue *) malloc(sizeof(queue));
root = (tree *) malloc(sizeof(tree));
node = root;
if (item == -1)
return;
node->data = item;
node->left = NULL;
node->right = NULL;
enque(&Q, node);
do {
dequeue(&Q, &ret);

if (left == -1)
break;
temp_l = (tree *) malloc(sizeof(tree));
temp_l->data = left;
temp_l->left = NULL;
temp_l->right = NULL;
ret->left = temp_l;
enque(&Q, temp_l);

if (right == -1)
break;
temp_r = (tree *) malloc(sizeof(tree));
temp_r->data = right;
temp_r->left = NULL;
temp_r->right = NULL;
ret->right = temp_r;
enque(&Q, temp_r);
} while (1);
printf(“calling inorder\n”);
inorderTraversal(&root);
}

int main() {
tree * root = NULL;
return 0;
}

Can we simply do this?

struct node* createTree(int m)
{
if(m<=size)
{
struct node *q = createNode(arr[m]);
q->left = createTree(m*2);
q->right = createTree(2*m + 1);
return q;
}
else return NULL;

• Meghasyam

Hi Ravi

why so complex solution ,
we can simply do in 2 steps

1. traverse the list and get one node at a time
2. send this value to complete binary tree functions

/* Paste your code here (You may delete these lines if not writing code) */

• Uddhav

What is the time complexity in case of your solution? I think it will be nlogn, whereas we find here the complexity to be n.

• Ravi Chandra

Hi Meghasyam,

You have simplified the second step to one sentence. When it actually comes to implementation, you will need the steps that I have given.

Can you elaborate step 2 on how are you going to implement?

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• Uddhav

This is nice, much like a BFS traversal for tree or say it is reversing BFS traversal and creating BST out of it ..