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Construct Binary Tree from given Parent Array representation | Iterative Approach
• Difficulty Level : Hard
• Last Updated : 30 Jun, 2020

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ implementation of the approach``#include ``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;``}`

## Java

 `// 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 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 constructed tree\n"``); ``    ``inorder(root); `` ` `} ``}`` ` `// This code is contributed by Arnab Kundu`

## Python3

 `# Python implementation of the approach`` ` `# A tree node``class` `Node:``    ``def` `__init__(``self``):``        ``self``.key ``=` `0``        ``self``.left ``=` `None``        ``self``.right ``=` `None`` ` `# Utility function to create new Node``def` `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 tree``def` `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 array``def` `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 Code``if` `__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#

 `// 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 constructed tree\n"``); ``    ``inorder(root); ``} ``}`` ` `// This code is contributed by Rajput-Ji`
Output:
```Inorder Traversal of constructed tree
6 5 3 1 4 0 2
```

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