Construct Binary Tree from given Parent Array representation | Iterative Approach
Last Updated :
01 Feb, 2023
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++
// 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;
}
// 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 constructed tree\n");
inorder(root);
}
}
// This code is contributed by Arnab Kundu
# 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# 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
<script>
// javascript implementation of the approach
// A tree node
class 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>
Output: Inorder Traversal of constructed tree
6 5 3 1 4 0 2
Time complexity: O(n) where n is the number of elements in the array.
Space Complexity: O(n), where n is the number of elements in the array.
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