Find the parent of a node in the given binary tree

Given a tree and a node, the task is to find the parent of the given node in the tree. Print -1 if the given node is the root node.
Examples:

Input: Node = 3
     1
   /   \
  2     3
 / \
4   5
Output: 1

Input: Node = 1
     1
   /   \
  2     3
 /       \
4         5
         /
        6
Output: -1

Approach: Write a recursive function that takes the current node and its parent as the arguments (root node is passed with -1 as its parent). If the current node is equal to the required node then print its parent and return else call the function recursively for its children and the current node as the parent.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <iostream>
using namespace std;

/* A binary tree node has data, pointer
to left child and a pointer
to right child */
struct Node {
    int data;
    struct Node *left, *right;
    Node(int data)
    {
        this->data = data;
        left = right = NULL;
    }
};

// Recursive function to find the
// parent of the given node
void findParent(struct Node* node,
                int val, int parent)
{
    if (node == NULL)
        return;

    // If current node is the required node
    if (node->data == val) {

        // Print its parent
        cout << parent;
    }
    else {

        // Recursive calls for the children
        // of the current node
        // Current node is now the new parent
        findParent(node->left, val, node->data);
        findParent(node->right, val, node->data);
    }
}

// Driver code
int main()
{
    struct Node* root = new Node(1);
    root->left = new Node(2);
    root->right = new Node(3);
    root->left->left = new Node(4);
    root->left->right = new Node(5);
    int node = 3;

    findParent(root, node, -1);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

/* A binary tree node has data, pointer
to left child and a pointer
to right child */
static class Node
{
    int data;
    Node left, right;
    Node(int data)
    {
        this.data = data;
        left = right = null;
    }
};

// Recursive function to find the
// parent of the given node
static void findParent(Node node,
                       int val, int parent)
{
    if (node == null)
        return;

    // If current node is the required node
    if (node.data == val)
    {

        // Print its parent
        System.out.print(parent);
    }
    else
    {

        // Recursive calls for the children
        // of the current node
        // Current node is now the new parent
        findParent(node.left, val, node.data);
        findParent(node.right, val, node.data);
    }
}

// Driver code
public static void main(String []args)
{
    Node root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(3);
    root.left.left = new Node(4);
    root.left.right = new Node(5);
    int node = 3;

    findParent(root, node, -1);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of
# the above approach

''' A binary tree node has data, pointer
to left child and a pointer
to right child '''
class Node:

    def __init__(self, data):

        self.data = data
        self.left = None
        self.right = None

# Recursive function to find the
# parent of the given node
def findParent(node : Node,
               val : int,
               parent : int) -> None:
    if (node is None):
        return

    # If current node is
    # the required node
    if (node.data == val):

        # Print its parent
        print(parent)
    else:

        # Recursive calls
        # for the children
        # of the current node
        # Current node is now
        # the new parent
        findParent(node.left,
                   val, node.data)
        findParent(node.right,
                   val, node.data)

# Driver code
if __name__ == "__main__":

    root = Node(1)
    root.left = Node(2)
    root.right = Node(3)
    root.left.left = Node(4)
    root.left.right = Node(5)
    node = 3
    findParent(root, node, -1)

# This code is contributed by sanjeev2552

C#

// C# implementation of the approach
using System;

class GFG
{

/* A binary tree node has data, pointer
to left child and a pointer
to right child */
public class Node
{
    public int data;
    public Node left, right;
    public Node(int data)
    {
        this.data = data;
        left = right = null;
    }
};

// Recursive function to find the
// parent of the given node
static void findParent(Node node,
                         int val, int parent)
{
    if (node == null)
        return;

    // If current node is the required node
    if (node.data == val)
    {

        // Print its parent
        Console.Write(parent);
    }
    else
    {

        // Recursive calls for the children
        // of the current node
        // Current node is now the new parent
        findParent(node.left, val, node.data);
        findParent(node.right, val, node.data);
    }
}

// Driver code
public static void Main(String []args)
{
    Node root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(3);
    root.left.left = new Node(4);
    root.left.right = new Node(5);
    int node = 3;

    findParent(root, node, -1);
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// JavaScript implementation of the approach

/* A binary tree node has data, pointer
to left child and a pointer
to right child */
class Node
{
    constructor(data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
};

// Recursive function to find the
// parent of the given node
function findParent(node, val, parent)
{
    if (node == null)
        return;

    // If current node is the required node
    if (node.data == val)
    {

        // Print its parent
        document.write(parent);
    }
    else
    {

        // Recursive calls for the children
        // of the current node
        // Current node is now the new parent
        findParent(node.left, val, node.data);
        findParent(node.right, val, node.data);
    }
}

// Driver code
var root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
var node = 3;

findParent(root, node, -1);


</script>

Output:

1

Time Complexity: O(N).
Auxiliary Space: O(N).