Construct Binary Tree from given Parent Array representation | Iterative Approach

Given an array that represents a tree in such a way that array indexes are values in tree nodes and array values give the parent node of that particular index (or node). The value of the root node index would always be -1 as there is no parent for root. Construct the standard linked representation of given Binary Tree from this given representation.

Examples:

Input: parent[] = {1, 5, 5, 2, 2, -1, 3}
Output:
Inorder Traversal of constructed tree
0 1 5 6 3 2 4
          5
        /  \
       1    2
      /    / \
     0    3   4
         /
        6 
Index of -1 is 5. So 5 is root.  
5 is present at indexes 1 and 2.  So 1 and 2 are
children of 5.  
1 is present at index 0, so 0 is child of 1.
2 is present at indexes 3 and 4.  So 3 and 4 are
children of 2.  
3 is present at index 6, so 6 is child of 3.

Input: parent[] = {-1, 0, 0, 1, 1, 3, 5}
Output:
Inorder Traversal of constructed tree
6 5 3 1 4 0 2
         0
       /   \
      1     2
     / \
    3   4
   /
  5 
 /
6

Approach: Recursive approach to this problem is discussed here.
Following is the iterative approach:



1. Create a map with key as the array index and its 
   value as the node for that index.
2. Start traversing the given parent array.
3. For all elements of the given array:
   (a) Search the map for the current index.
       (i) If the current index does not exist in the map:
           .. Create a node for the current index
           .. Map the newly created node with its key by m[i]=node
       (ii) If the key exists in the map:
            .. it means that the node is already created
            .. Do nothing
   (b) If the parent of the current index is -1, it implies it is
       the root of the tree
       .. Make root=m[i]
       Else search for the parent in the map
       (i) If the parent does not exist:
           .. Create the parent node.
           .. Assign the current node as its left child 
           .. Map the parent node(as in Step 3.(a).(i))
       (ii) If the parent exists:
           .. If the left child of the parent does not exist
              -> Assign the node as its left child
           .. Else (i.e. right child of the parent does not exist)
              -> Assign the node as its right child

This approach works even when the nodes are not given in order.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// A tree node
struct Node {
    int key;
    struct Node *left, *right;
};
  
// Utility function to create new Node
Node* newNode(int key)
{
    Node* temp = new Node;
    temp->key = key;
    temp->left = temp->right = NULL;
    return (temp);
}
  
// Utility function to perform
// inorder traversal of the tree
void inorder(Node* root)
{
    if (root != NULL) {
        inorder(root->left);
        cout << root->key << " ";
        inorder(root->right);
    }
}
  
// Function to construct a Binary Tree from parent array
Node* createTree(int parent[], int n)
{
    // A map to keep track of all the nodes created.
    // Key: node value; Value: Pointer to that Node
    map<int, Node*> m;
    Node *root, *temp;
    int i;
  
    // Iterate for all elements of the parent array.
    for (i = 0; i < n; i++) {
  
        // Node i does not exist in the map
        if (m.find(i) == m.end()) {
  
            // Create a new node for the current index
            temp = newNode(i);
  
            // Entry of the node in the map with
            // key as i and value as temp
            m[i] = temp;
        }
  
        // If parent is -1
        // Current node i is the root
        // So mark it as the root of the tree
        if (parent[i] == -1)
            root = m[i];
  
        // Current node is not root and parent
        // of that node is not created yet
        else if (m.find(parent[i]) == m.end()) {
  
            // Create the parent
            temp = newNode(parent[i]);
  
            // Assign the node as the
            // left child of the parent
            temp->left = m[i];
  
            // Entry of parent in map
            m[parent[i]] = temp;
        }
  
        // Current node is not root and parent
        // of that node is already created
        else {
  
            // Left child of the parent doesn't exist
            if (!m[parent[i]]->left)
                m[parent[i]]->left = m[i];
  
            // Right child of the parent doesn't exist
            else
                m[parent[i]]->right = m[i];
        }
    }
    return root;
}
  
// Driver code
int main()
{
    int parent[] = { -1, 0, 0, 1, 1, 3, 5 };
    int n = sizeof parent / sizeof parent[0];
    Node* root = createTree(parent, n);
    cout << "Inorder Traversal of constructed tree\n";
    inorder(root);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach 
import java.util.*;
  
class GFG
{
      
// A tree node 
static class Node 
    int key; 
    Node left, right; 
}; 
  
// Utility function to create new Node 
static Node newNode(int key) 
    Node temp = new Node(); 
    temp.key = key; 
    temp.left = temp.right = null
    return (temp); 
  
// Utility function to perform 
// inorder traversal of the tree 
static void inorder(Node root) 
    if (root != null
    
        inorder(root.left); 
        System.out.print( root.key + " "); 
        inorder(root.right); 
    
  
// Function to construct a Binary Tree from parent array 
static Node createTree(int parent[], int n) 
    // A map to keep track of all the nodes created. 
    // Key: node value; Value: Pointer to that Node 
    HashMap<Integer, Node> m=new HashMap<>(); 
    Node root=new Node(), temp=new Node(); 
    int i; 
  
    // Iterate for all elements of the parent array. 
    for (i = 0; i < n; i++)
    
  
        // Node i does not exist in the map 
        if (m.get(i) == null
        
  
            // Create a new node for the current index 
            temp = newNode(i); 
  
            // Entry of the node in the map with 
            // key as i and value as temp 
            m.put(i, temp); 
        
  
        // If parent is -1 
        // Current node i is the root 
        // So mark it as the root of the tree 
        if (parent[i] == -1
            root = m.get(i); 
  
        // Current node is not root and parent 
        // of that node is not created yet 
        else if (m.get(parent[i]) == null
        
  
            // Create the parent 
            temp = newNode(parent[i]); 
  
            // Assign the node as the 
            // left child of the parent 
            temp.left = m.get(i); 
  
            // Entry of parent in map 
            m.put(parent[i],temp); 
        
  
        // Current node is not root and parent 
        // of that node is already created 
        else
        
  
            // Left child of the parent doesn't exist 
            if (m.get(parent[i]).left == null
                m.get(parent[i]).left = m.get(i); 
  
            // Right child of the parent doesn't exist 
            else
                m.get(parent[i]).right = m.get(i); 
        
    
    return root; 
  
// Driver code 
public static void main(String args[]) 
    int parent[] = { -1, 0, 0, 1, 1, 3, 5 }; 
    int n = parent.length; 
    Node root = createTree(parent, n); 
    System.out.print( "Inorder Traversal of coned tree\n"); 
    inorder(root); 
  
}
  
// This code is contributed by Arnab Kundu

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
using System.Collections.Generic;
  
class GFG
{
      
// A tree node 
class Node 
    public int key; 
    public Node left, right; 
}; 
  
// Utility function to create new Node 
static Node newNode(int key) 
    Node temp = new Node(); 
    temp.key = key; 
    temp.left = temp.right = null
    return (temp); 
  
// Utility function to perform 
// inorder traversal of the tree 
static void inorder(Node root) 
    if (root != null
    
        inorder(root.left); 
        Console.Write( root.key + " "); 
        inorder(root.right); 
    
  
// Function to construct a Binary Tree from parent array 
static Node createTree(int []parent, int n) 
    // A map to keep track of all the nodes created. 
    // Key: node value; Value: Pointer to that Node 
    Dictionary<int, Node> m = new Dictionary<int, Node>();
    Node root = new Node(), temp = new Node(); 
    int i; 
  
    // Iterate for all elements of the parent array. 
    for (i = 0; i < n; i++)
    
  
        // Node i does not exist in the map 
        if (!m.ContainsKey(i)) 
        
  
            // Create a new node for the current index 
            temp = newNode(i); 
  
            // Entry of the node in the map with 
            // key as i and value as temp 
            m.Add(i, temp); 
        
  
        // If parent is -1 
        // Current node i is the root 
        // So mark it as the root of the tree 
        if (parent[i] == -1) 
            root = m[i]; 
  
        // Current node is not root and parent 
        // of that node is not created yet 
        else if (!m.ContainsKey(parent[i])) 
        
  
            // Create the parent 
            temp = newNode(parent[i]); 
  
            // Assign the node as the 
            // left child of the parent 
            temp.left = m[i]; 
  
            // Entry of parent in map 
            m.Add(parent[i], temp); 
        
  
        // Current node is not root and parent 
        // of that node is already created 
        else
        
  
            // Left child of the parent doesn't exist 
            if (m[parent[i]].left == null
                m[parent[i]].left = m[i]; 
  
            // Right child of the parent doesn't exist 
            else
                m[parent[i]].right = m[i]; 
        
    
    return root; 
  
// Driver code 
public static void Main(String []args) 
    int []parent = { -1, 0, 0, 1, 1, 3, 5 }; 
    int n = parent.Length; 
    Node root = createTree(parent, n); 
    Console.Write("Inorder Traversal of coned tree\n"); 
    inorder(root); 
}
  
// This code is contributed by Rajput-Ji

chevron_right


Output:

Inorder Traversal of constructed tree
6 5 3 1 4 0 2


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks 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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : andrew1234, Rajput-Ji