Print the nodes having exactly one child in a Binary tree

Given a binary tree, the task is to print all the nodes having exactly one child. Print “-1” if no such node exists.
Examples:

Input: 
            2
           / \
          3   5
         /   / \
        7   8   6
Output: 3
Explanation:
There is only one node having
single child that is 3.

Input:
           9
          / \
         7   8
            / \
           4   3
Output: -1
Explanation:
There is no node having exactly one
child in the binary tree. 

Approach: The idea is to traverse the tree in the inorder traversal and at each step of the traversal check that if the node is having exactly one child. Then append that node into a result array to keep track of such nodes. After the traversal, simply print each element of this result array.

Below is the implementation of the above approach: 

C++

// C++ implementation to print
// the nodes having a single child
#include <bits/stdc++.h>
using namespace std;

// Class of the Binary Tree node
struct Node
{
  int data;
  Node *left, *right;

  Node(int x)
  {
    data = x;
    left = right = NULL;
  }
};

vector<int> lst;

// Function to find the nodes
// having single child
void printNodesOneChild(Node* root)
{
  // Base case
  if (root == NULL)
    return;

  // Condition to check if the
  // node is having only one child
  if (root->left != NULL &&
      root->right == NULL ||
      root->left == NULL &&
      root->right != NULL)
  {
    lst.push_back(root->data);
  }

  // Traversing the left child
  printNodesOneChild(root -> left);

  // Traversing the right child
  printNodesOneChild(root -> right);
}

//Driver code
int main()
{
  // Constructing the binary tree
  Node *root = new Node(2);
  root -> left = new Node(3);
  root -> right = new Node(5);
  root -> left -> left = new Node(7);
  root -> right -> left = new Node(8);
  root -> right -> right = new Node(6);

  // Function calling
  printNodesOneChild(root);

  // Condition to check if there is
  // no such node having single child
  if (lst.size() == 0)
    printf("-1");
  else
  {
    for(int value : lst)
    {
      cout << (value) << endl;
    }
  }
}

// This code is contributed by Mohit Kumar 29

Java

// Java implementation to print
// the nodes having a single child
import java.util.Vector;

// Class of the Binary Tree node
class Node
{
    int data;
    Node left, right;

    Node(int data)
    {
        this.data = data;
    }
}

class GFG{

// List to keep track of nodes
// having single child
static Vector<Integer> list = new Vector<>();

// Function to find the nodes
// having single child
static void printNodesOneChild(Node root)
{
    // Base case
    if (root == null)
        return;

    // Condition to check if the node
    // is having only one child
    if (root.left != null && root.right == null ||
        root.left == null && root.right != null)
    {
        list.add(root.data);
    }

    // Traversing the left child
    printNodesOneChild(root.left);

    // Traversing the right child
    printNodesOneChild(root.right);
}

// Driver code
public static void main(String[] args)
{

    // Constructing the binary tree
    Node root = new Node(2);
    root.left = new Node(3);
    root.right = new Node(5);
    root.left.left = new Node(7);
    root.right.left = new Node(8);
    root.right.right = new Node(6);

    // Function calling
    printNodesOneChild(root);

    // Condition to check if there is
    // no such node having single child
    if (list.size() == 0)
        System.out.println(-1);
    else
    {
        for(int value : list)
        {
            System.out.println(value);
        }
    }
}
}

// This code is contributed by sumit_9

Python3

# Python3 implementation to print
# the nodes having a single child

# Class of the Binary Tree node
class node:

    # Constructor to construct
    # the node of the Binary tree
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

# List to keep track of
# nodes having single child
single_child_nodes = []

# Function to find the nodes
# having single child
def printNodesOneChild(root):

    # Base Case
    if not root:
        return

    # Condition to check if the node
    # is having only one child
    if not root.left and root.right:
        single_child_nodes.append(root)
    elif root.left and not root.right:
        single_child_nodes.append(root)

    # Traversing the left child
    printNodesOneChild(root.left)

    # Traversing the right child
    printNodesOneChild(root.right)
    return

# Driver Code
if __name__ == "__main__":

    # Construction of Binary Tree
    root = node(2)
    root.left = node(3)
    root.right = node(5)
    root.left.left = node(7)
    root.right.left = node(8)
    root.right.right = node(6)

    # Function Call
    printNodesOneChild(root)

    # Condition to check if there is
    # no such node having single child
    if not len(single_child_nodes):
        print(-1)
    else:
        for i in single_child_nodes:
            print(i.val, end = " ")
        print()

C#

// C# implementation to print
// the nodes having a single child
using System;
using System.Collections;

class GFG{

// Structure of binary tree
public class Node
{
    public Node left;
    public Node right;
    public int data;
};

// Function to create a new node
static Node newNode(int key)
{
    Node node = new Node();
    node.left = node.right = null;
    node.data = key;
    return node;
}

// List to keep track of nodes
// having single child
static ArrayList list = new ArrayList();

// Function to find the nodes
// having single child
static void printNodesOneChild(Node root)
{

    // Base case
    if (root == null)
        return;

    // Condition to check if the node
    // is having only one child
    if (root.left != null && root.right == null ||
        root.left == null && root.right != null)
    {
        list.Add(root.data);
    }

    // Traversing the left child
    printNodesOneChild(root.left);

    // Traversing the right child
    printNodesOneChild(root.right);
}

// Driver code
static public void Main()
{

    // Constructing the binary tree
    Node root = newNode(2);
    root.left = newNode(3);
    root.right = newNode(5);
    root.left.left = newNode(7);
    root.right.left = newNode(8);
    root.right.right = newNode(6);

    // Function calling
    printNodesOneChild(root);

    // Condition to check if there is
    // no such node having single child
    if (list.Count == 0)
        Console.WriteLine(-1);
    else
    {
        foreach(int value in list)
        {
            Console.WriteLine(value);
        }
    }
}
}

// This code is contributed by offbeat

Javascript

<script>
// Javascript implementation to print
// the nodes having a single child

// Class of the Binary Tree node
class Node
{
    constructor(data)
    {
        this.data = data;
        this.left=this.right=null;
    }

}

// List to keep track of nodes
// having single child
let list = [];

// Function to find the nodes
// having single child
function printNodesOneChild(root)
{
    // Base case
    if (root == null)
        return;

    // Condition to check if the node
    // is having only one child
    if (root.left != null && root.right == null ||
        root.left == null && root.right != null)
    {
        list.push(root.data);
    }

    // Traversing the left child
    printNodesOneChild(root.left);

    // Traversing the right child
    printNodesOneChild(root.right);
}

// Driver code
// Constructing the binary tree
    let root = new Node(2);
    root.left = new Node(3);
    root.right = new Node(5);
    root.left.left = new Node(7);
    root.right.left = new Node(8);
    root.right.right = new Node(6);

    // Function calling
    printNodesOneChild(root);

    // Condition to check if there is
    // no such node having single child
    if (list.length == 0)
        document.write(-1);
    else
    {
        document.write(list.join("<br>"))
    }

// This code is contributed by patel2127
</script>

Output

3

Time complexity: O(n) where n is no of nodes in binary tree

Auxiliary Space: O(n)