# Construct Binary Tree from given Parent Array representation

• Difficulty Level : Hard
• Last Updated : 28 Jul, 2022

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: root of below tree
5
/  \
1    2
/    / \
0    3   4
/
6
Explanation:
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: root of below tree
0
/   \
1     2
/ \
3   4
/
5
/
6```

Expected time complexity is O(n) where n is number of elements in given array.

We strongly recommend to minimize your browser and try this yourself first.

A Simple Solution to recursively construct by first searching the current root, then recurring for the found indexes (there can be at most two indexes) and making them left and right subtrees of root. This solution takes O(n2) as we have to linearly search for every node.

An Efficient Solution can solve the above problem in O(n) time. The idea is to use extra space. An array created[0..n-1] is used to keep track of created nodes.
createTree(parent[], n)

1. Create an array of pointers say created[0..n-1]. The value of created[i] is NULL if node for index i is not created, else value is pointer to the created node.
2. Do following for every index i of given array
createNode(parent, i, created)

createNode(parent[], i, created[])

1. If created[i] is not NULL, then node is already created. So return.
2. Create a new node with value ‘i’.
3. If parent[i] is -1 (i is root), make created node as root and return.
4. Check if parent of ‘i’ is created (We can check this by checking if created[parent[i]] is NULL or not.
5. If parent is not created, recur for parent and create the parent first.
6. Let the pointer to parent be p. If p->left is NULL, then make the new node as left child. Else make the new node as right child of parent.

Following is C++ implementation of above idea.

## C++14

 `// C++ program to construct a Binary Tree from parent array``#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);``}` `// Creates a node with key as 'i'.  If i is root, then it changes``// root.  If parent of i is not created, then it creates parent first``void` `createNode(``int` `parent[], ``int` `i, Node *created[], Node **root)``{``    ``// If this node is already created``    ``if` `(created[i] != NULL)``        ``return``;` `    ``// Create a new node and set created[i]``    ``created[i] = newNode(i);` `    ``// If 'i' is root, change root pointer and return``    ``if` `(parent[i] == -1)``    ``{``        ``*root = created[i];``        ``return``;``    ``}` `    ``// If parent is not created, then create parent first``    ``if` `(created[parent[i]] == NULL)``        ``createNode(parent, parent[i], created, root);` `    ``// Find parent pointer``    ``Node *p = created[parent[i]];` `    ``// If this is first child of parent``    ``if` `(p->left == NULL)``        ``p->left = created[i];``    ``else` `// If second child``        ``p->right = created[i];``}` `// Creates tree from parent[0..n-1] and returns root of the created tree``Node *createTree(``int` `parent[], ``int` `n)``{``    ``// Create an array created[] to keep track``    ``// of created nodes, initialize all entries``    ``// as NULL``    ``Node *created[n];``    ``for` `(``int` `i=0; ileft);``        ``cout << root->key << ``" "``;``        ``inorder(root->right);``    ``}``}` `// Driver method``int` `main()``{``    ``int` `parent[] =  {-1, 0, 0, 1, 1, 3, 5};``    ``int` `n = ``sizeof` `parent / ``sizeof` `parent;``    ``Node *root = createTree(parent, n);``    ``cout << ``"Inorder Traversal of constructed tree\n"``;``    ``inorder(root);``    ``newLine();``}`

## Java

 `// Java program to construct a binary tree from parent array`` ` `// A binary tree node``class` `Node``{``    ``int` `key;``    ``Node left, right;`` ` `    ``public` `Node(``int` `key)``    ``{``        ``this``.key = key;``        ``left = right = ``null``;``    ``}``}`` ` `class` `BinaryTree``{``    ``Node root;`` ` `    ``// Creates a node with key as 'i'.  If i is root, then it changes``    ``// root.  If parent of i is not created, then it creates parent first``    ``void` `createNode(``int` `parent[], ``int` `i, Node created[])``    ``{``        ``// If this node is already created``        ``if` `(created[i] != ``null``)``            ``return``;`` ` `        ``// Create a new node and set created[i]``        ``created[i] = ``new` `Node(i);`` ` `        ``// If 'i' is root, change root pointer and return``        ``if` `(parent[i] == -``1``)``        ``{``            ``root = created[i];``            ``return``;``        ``}`` ` `        ``// If parent is not created, then create parent first``        ``if` `(created[parent[i]] == ``null``)``            ``createNode(parent, parent[i], created);`` ` `        ``// Find parent pointer``        ``Node p = created[parent[i]];`` ` `        ``// If this is first child of parent``        ``if` `(p.left == ``null``)``            ``p.left = created[i];``        ``else` `// If second child``         ` `            ``p.right = created[i];``    ``}`` ` `    ``/* Creates tree from parent[0..n-1] and returns root of``       ``the created tree */``    ``Node createTree(``int` `parent[], ``int` `n)``    ``{   ``        ``// Create an array created[] to keep track``        ``// of created nodes, initialize all entries``        ``// as NULL``        ``Node[] created = ``new` `Node[n];``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``created[i] = ``null``;`` ` `        ``for` `(``int` `i = ``0``; i < n; i++)``            ``createNode(parent, i, created);`` ` `        ``return` `root;``    ``}`` ` `    ``//For adding new line in a program``    ``void` `newLine()``    ``{``        ``System.out.println(``""``);``    ``}`` ` `    ``// Utility function to do inorder traversal``    ``void` `inorder(Node node)``    ``{``        ``if` `(node != ``null``)``        ``{``            ``inorder(node.left);``            ``System.out.print(node.key + ``" "``);``            ``inorder(node.right);``        ``}``    ``}`` ` `    ``// Driver method``    ``public` `static` `void` `main(String[] args)``    ``{`` ` `        ``BinaryTree tree = ``new` `BinaryTree();``        ``int` `parent[] = ``new` `int``[]{-``1``, ``0``, ``0``, ``1``, ``1``, ``3``, ``5``};``        ``int` `n = parent.length;``        ``Node node = tree.createTree(parent, n);``        ``System.out.println(``"Inorder traversal of constructed tree "``);``        ``tree.inorder(node);``        ``tree.newLine();``    ``}``}`` ` `// This code has been contributed by Mayank Jaiswal(mayank_24)`

## Python3

 `# Python implementation to construct a Binary Tree from``# parent array` `# A node structure``class` `Node:``    ``# A utility function to create a new node``    ``def` `__init__(``self``, key):``        ``self``.key ``=` `key``        ``self``.left ``=` `None``        ``self``.right ``=` `None` `""" Creates a node with key as 'i'. If i is root,then``    ``it changes root. If parent of i is not created, then``    ``it creates parent first``"""``def` `createNode(parent, i, created, root):` `    ``# If this node is already created``    ``if` `created[i] ``is` `not` `None``:``        ``return` `    ``# Create a new node and set created[i]``    ``created[i] ``=` `Node(i)` `    ``# If 'i' is root, change root pointer and return``    ``if` `parent[i] ``=``=` `-``1``:``        ``root[``0``] ``=` `created[i] ``# root denotes root of the tree``        ``return` `    ``# If parent is not created, then create parent first``    ``if` `created[parent[i]] ``is` `None``:``        ``createNode(parent, parent[i], created, root )` `    ``# Find parent pointer``    ``p ``=` `created[parent[i]]` `    ``# If this is first child of parent``    ``if` `p.left ``is` `None``:``        ``p.left ``=` `created[i]``    ``# If second child``    ``else``:``        ``p.right ``=` `created[i]`  `# Creates tree from parent[0..n-1] and returns root of the``# created tree``def` `createTree(parent):``    ``n ``=` `len``(parent)``    ` `    ``# Create and array created[] to keep track``    ``# of created nodes, initialize all entries as None``    ``created ``=` `[``None` `for` `i ``in` `range``(n``+``1``)]``    ` `    ``root ``=` `[``None``]``    ``for` `i ``in` `range``(n):``        ``createNode(parent, i, created, root)` `    ``return` `root[``0``]` `#Inorder traversal of tree``def` `inorder(root):``    ``if` `root ``is` `not` `None``:``        ``inorder(root.left)``        ``print` `(root.key,end``=``" "``)``        ``inorder(root.right)` `# Driver Method``parent ``=` `[``-``1``, ``0``, ``0``, ``1``, ``1``, ``3``, ``5``]``root ``=` `createTree(parent)``print` `(``"Inorder Traversal of constructed tree"``)``inorder(root)` `# This code is contributed by Nikhil Kumar Singh(nickzuck_007)`

## C#

 `// C# program to construct a binary``// tree from parent array``using` `System;` `// A binary tree node``public` `class` `Node``{``    ``public` `int` `key;``    ``public` `Node left, right;` `    ``public` `Node(``int` `key)``    ``{``        ``this``.key = key;``        ``left = right = ``null``;``    ``}``}` `class` `GFG``{``public` `Node root;` `// Creates a node with key as 'i'.``// If i is root, then it changes``// root. If parent of i is not created,``// then it creates parent first``public` `virtual` `void` `createNode(``int``[] parent,``                               ``int` `i, Node[] created)``{``    ``// If this node is already created``    ``if` `(created[i] != ``null``)``    ``{``        ``return``;``    ``}` `    ``// Create a new node and set created[i]``    ``created[i] = ``new` `Node(i);` `    ``// If 'i' is root, change root``    ``// pointer and return``    ``if` `(parent[i] == -1)``    ``{``        ``root = created[i];``        ``return``;``    ``}` `    ``// If parent is not created, then``    ``// create parent first``    ``if` `(created[parent[i]] == ``null``)``    ``{``        ``createNode(parent, parent[i], created);``    ``}` `    ``// Find parent pointer``    ``Node p = created[parent[i]];` `    ``// If this is first child of parent``    ``if` `(p.left == ``null``)``    ``{``        ``p.left = created[i];``    ``}``    ``else` `// If second child``    ``{` `        ``p.right = created[i];``    ``}``}` `/* Creates tree from parent[0..n-1]``and returns root of the created tree */``public` `virtual` `Node createTree(``int``[] parent, ``int` `n)``{``    ``// Create an array created[] to``    ``// keep track of created nodes,``    ``// initialize all entries as NULL``    ``Node[] created = ``new` `Node[n];``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``created[i] = ``null``;``    ``}` `    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``createNode(parent, i, created);``    ``}` `    ``return` `root;``}` `// For adding new line in a program``public` `virtual` `void` `newLine()``{``    ``Console.WriteLine(``""``);``}` `// Utility function to do inorder traversal``public` `virtual` `void` `inorder(Node node)``{``    ``if` `(node != ``null``)``    ``{``        ``inorder(node.left);``        ``Console.Write(node.key + ``" "``);``        ``inorder(node.right);``    ``}``}` `// Driver Code``public` `static` `void` `Main(``string``[] args)``{``    ``GFG tree = ``new` `GFG();``    ``int``[] parent = ``new` `int``[]{-1, 0, 0, 1, 1, 3, 5};``    ``int` `n = parent.Length;``    ``Node node = tree.createTree(parent, n);``    ``Console.WriteLine(``"Inorder traversal of "` `+``                          ``"constructed tree "``);``    ``tree.inorder(node);``    ``tree.newLine();``}``}` `// This code is contributed by Shrikant13`

## Javascript

 ``

Output

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

Another Efficient Solution:

The idea is to first create all n new tree nodes, each having values from 0 to n – 1, where n is the size of parent array, and store them in any data structure like map, array etc to keep track of which node is created for which value. Then traverse the given parent array and build the tree by setting the parent-child relationship.

Following is the C++ implementation of the above idea.

## C++

 `// C++ program to construct a Binary Tree from parent array``#include ``using` `namespace` `std;`  `struct` `Node {``    ``int` `data;``    ``struct` `Node* left = NULL;``    ``struct` `Node* right = NULL;``    ``Node() {}` `    ``Node(``int` `x) { data = x; }``};` `// Function to construct binary tree from parent array.``Node* createTree(``int` `parent[], ``int` `n)``{``    ``// Create an array to store the reference``    ``// of all newly created nodes corresponding``    ``// to node value``    ``vector ref;` `    ``// This root represent the root of the``    ``// newly constructed tree``    ``Node* root = ``new` `Node();` `    ``// Create n new tree nodes, each having``    ``// a value from 0 to n-1, and store them``    ``// in ref``    ``for` `(``int` `i = 0; i < n; i++) {``        ``Node* temp = ``new` `Node(i);``        ``ref.push_back(temp);``    ``}` `    ``// Traverse the parent array and build the tree``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// If the parent is -1, set the root``        ``// to the current node having``        ``// the value i which is stored in ref[i]``        ``if` `(parent[i] == -1) {``            ``root = ref[i];``        ``}``        ``else` `{``            ``// Check if the parent's left child``            ``// is NULL then map the left child``            ``// to its parent.``            ``if` `(ref[parent[i]]->left == NULL)``                ``ref[parent[i]]->left = ref[i];``            ``else``                ``ref[parent[i]]->right = ref[i];``        ``}``    ``}` `    ``// Return the root of the newly constructed tree``    ``return` `root;``}` `// Function for inorder traversal``void` `inorder(Node* root)``{``    ``if` `(root != NULL) {``        ``inorder(root->left);``        ``cout << root->data << ``" "``;``        ``inorder(root->right);``    ``}``}` `// Driver code``int` `main()``{``    ``int` `parent[] = { -1, 0, 0, 1, 1, 3, 5 };``    ``int` `n = ``sizeof` `parent / ``sizeof` `parent;``    ``Node* root = createTree(parent, n);``    ``cout << ``"Inorder Traversal of constructed tree\n"``;``    ``inorder(root);``    ``return` `0;``}`

## Java

 `// Java program to construct a Binary Tree from parent array` `// A binary tree node``class` `Node {``    ``int` `key;``    ``Node left, right;``    ``public` `Node() {}``    ``public` `Node(``int` `key)``    ``{``        ``this``.key = key;``        ``left = right = ``null``;``    ``}``}` `class` `BinaryTree {``    ``Node root;` `    ``// Function to construct binary tree from parent array.``    ``Node createTree(``int` `parent[], ``int` `n)``    ``{``        ``// Create an array to store the reference``        ``// of all newly created nodes corresponding``        ``// to node value``        ``Node[] ref = ``new` `Node[n];` `        ``// This root represent the root of the``        ``// newly constructed tree``        ``Node root = ``new` `Node();` `        ``// Create n new tree nodes, each having``        ``// a value from 0 to n-1, and store them``        ``// in ref``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``Node temp = ``new` `Node(i);``            ``ref[i] = temp;``        ``}` `        ``// Traverse the parent array and build the tree``        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``// If the parent is -1, set the root``            ``// to the current node having``            ``// the value i which is stored in ref[i]``            ``if` `(parent[i] == -``1``) {``                ``root = ref[i];``            ``}``            ``else` `{``                ``// Check if the parent's left child``                ``// is NULL then map the left child``                ``// to its parent.``                ``if` `(ref[parent[i]].left == ``null``)``                    ``ref[parent[i]].left = ref[i];``                ``else``                    ``ref[parent[i]].right = ref[i];``            ``}``        ``}` `        ``// Return the root of the newly constructed tree``        ``return` `root;``    ``}` `    ``// function for inorder traversal``    ``void` `inorder(Node node)``    ``{``        ``if` `(node != ``null``) {``            ``inorder(node.left);``            ``System.out.print(node.key + ``" "``);``            ``inorder(node.right);``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{` `        ``BinaryTree tree = ``new` `BinaryTree();``        ``int` `parent[] = ``new` `int``[] { -``1``, ``0``, ``0``, ``1``, ``1``, ``3``, ``5` `};``        ``int` `n = parent.length;``        ``Node node = tree.createTree(parent, n);``        ``System.out.println(``            ``"Inorder traversal of constructed tree "``);``        ``tree.inorder(node);``    ``}``}` `// This code has been contributed by Abhijeet Kumar(abhijeet19403)`

Output

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

Time Complexity: O(n), where n is the size of parent array
Auxiliary Space: O(n)

Similar Problem: Find Height of Binary Tree represented by Parent array
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