Construct Binary Tree from given Parent Array representation | Iterative Approach - GeeksforGeeks

02 Jun, 2021
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++
Java
Python3
C#
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; // A tree nodestruct Node {    int key;    struct Node *left, *right;}; // Utility function to create new NodeNode* 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 treevoid inorder(Node* root){    if (root != NULL) {        inorder(root->left);        cout << root->key << " ";        inorder(root->right);    }} // Function to construct a Binary Tree from parent arrayNode* 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 codeint 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;}
// Java implementation of the approachimport java.util.*; class GFG{     // A tree nodestatic class Node{    int key;    Node left, right;}; // Utility function to create new Nodestatic 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 treestatic 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 arraystatic 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 codepublic 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 constructed tree\n");    inorder(root); }} // This code is contributed by Arnab Kundu
# Python implementation of the approach # A tree nodeclass Node:    def __init__(self):        self.key = 0        self.left = None        self.right = None # Utility function to create new Nodedef newNode(key: int) -> Node:    temp = Node()    temp.key = key    temp.left = None    temp.right = None    return temp # Utility function to perform# inorder traversal of the treedef inorder(root: Node):    if root is not None:        inorder(root.left)        print(root.key, end=" ")        inorder(root.right) # Function to construct a Binary Tree from parent arraydef createTree(parent: list, n: int) -> Node:     # A map to keep track of all the nodes created.    # Key: node value; Value: Pointer to that Node    m = dict()    root = Node()     # Iterate for all elements of the parent array.    for i in range(n):         # Node i does not exist in the map        if i not in m:             # 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        elif parent[i] not in m:             # 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 is None:                m[parent[i]].left = m[i]             # Right child of the parent doesn't exist            else:                m[parent[i]].right = m[i]    return root # Driver Codeif __name__ == "__main__":    parent = [-1, 0, 0, 1, 1, 3, 5]    n = len(parent)    root = createTree(parent, n)    print("Inorder Traversal of constructed tree")    inorder(root) # This code is contributed by# sanjeev2552
// C# implementation of the approachusing System;using System.Collections.Generic; class GFG{     // A tree nodeclass Node{    public int key;    public Node left, right;}; // Utility function to create new Nodestatic 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 treestatic 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 arraystatic 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 codepublic 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 constructed tree\n");    inorder(root);}} // This code is contributed by Rajput-Ji
<script>// javascript implementation of the approach     // A tree nodeclass Node {    constructor(val) {        this.key = val;        this.left = null;        this.right = null;    }}     // Utility function to create new Node    function newNode(key) {var temp = new Node();        temp.key = key;        temp.left = temp.right = null;        return (temp);    }     // Utility function to perform    // inorder traversal of the tree    function inorder(root) {        if (root != null) {            inorder(root.left);            document.write(root.key + " ");            inorder(root.right);        }    }     // Function to construct a Binary Tree from parent array    function createTree(parent , n) {        // A map to keep track of all the nodes created.        // Key: node value; Value: Pointer to that Node        var m = new Map();var root = new Node(), temp = new Node();        var 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.set(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.set(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             var parent = [ -1, 0, 0, 1, 1, 3, 5 ];        var n = parent.length;var root = createTree(parent, n);        document.write("Inorder Traversal of constructed tree<br/>");        inorder(root); // This code contributed by umadevi9616</script>
Inorder Traversal of constructed tree
6 5 3 1 4 0 2
 
andrew1234
Rajput-Ji
sanjeev2552
Akanksha_Rai
umadevi9616
Binary Tree
Constructive Algorithms
Marketing
Arrays
Tree
Arrays
Tree
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Comments
Old Comments
Window Sliding Technique
Program to find sum of elements in a given array
Find duplicates in O(n) time and O(1) extra space | Set 1
Reversal algorithm for array rotation
Next Greater Element
Tree Traversals (Inorder, Preorder and Postorder)
Binary Tree | Set 1 (Introduction)
Level Order Binary Tree Traversal
AVL Tree | Set 1 (Insertion)
Inorder Tree Traversal without Recursion