Print the nodes that are just above the leaf node

Given a binary tree consisting of N nodes, the task is to print the nodes that are just above the leaf node.
Examples:

Input: N = 7, Below is the given Binary Tree: 

Output: 20 8 12 
Explanation:
Node 20 is just above the leaf node 22. 
Node 8 is just above the leaf node 4. 
Node 12 is just above the leaf nodes 10 and 14.

Input: N = 5, Below is the given Binary Tree: 

Output: 1 2 
Explanation:
Node 1 is just above the leaf node 3. 
Node 2 is just above the leaf nodes 4 and 5.

Approach: The idea is to traverse the tree and for each node, check if it can be the one that is just above the leaf node. For that, the current node must have children and at least one of the children should be a leaf node. Below are the steps:

• Traverse the tree And check each node.
• If the current node has two children, check if any of them is a root node. If yes, then print the current node.
• If the current node has only a left or right child, then check if that left or the right child is a leaf node. If yes, then print the current node.
• Else, continue traversing the tree and move to the next node.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Node of tree
struct node {
    int data;
    struct node *left, *right;
};

// Creates and initializes a new
// node for the tree
struct node* newnode(int data)
{
    struct node* temp = new node();
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}

// Prints all nodes which are just
// above leaf node
void cal(struct node* root)
{

    // If tree is empty
    if (root == NULL) {
        return;
    }

    // If it is a leaf node
    if (root->left == NULL
        && root->right == NULL) {
        return;
    }

    // For internal nodes
    else {

        // If node has two children
        if (root->left != NULL
            && root->right != NULL) {

            if ((root->left->left == NULL
                 && root->left->right == NULL)
                || (root->right->left == NULL
                    && root->right->right == NULL)) {

                cout << root->data << " ";
            }
        }

        // If node has only left child
        if (root->left != NULL
            && root->right == NULL) {

            if (root->left->left == NULL
                && root->left->right == NULL) {

                cout << root->data << " ";
            }
        }

        // If node has only right child
        if (root->right != NULL
            && root->left == NULL) {

            if (root->right->left == NULL
                && root->right->right == NULL) {

                cout << root->data << " ";
            }
        }
    }

    // Recursively Call for left
    // and right subtree
    cal(root->left);
    cal(root->right);
}

// Driver Code
int main()
{
    // Construct a tree
    node* root = newnode(20);
    root->left = newnode(8);
    root->right = newnode(22);
    root->left->left = newnode(4);
    root->left->right = newnode(12);
    root->left->right->left = newnode(10);
    root->left->right->right = newnode(14);

    // Function Call
    cal(root);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

// Class containing the left and right
// child of current node and the
// key value
class Node
{
    int data;
    Node left, right;

    // Constructor of the class
    public Node(int item)
    {
        data = item;
        left = right = null;
    }
}

class GFG{

Node root;

// Prints all nodes which are just
// above leaf node
static void cal(Node root)
{

    // If tree is empty
    if (root == null)
    {
        return;
    }

    // If it is a leaf node
    if (root.left == null &&
       root.right == null)
    {
        return;
    }

    // For internal nodes
    else
    {

        // If node has two children
        if (root.left != null &&
           root.right != null)
        {
            if ((root.left.left == null &&
                root.left.right == null) ||
               (root.right.left == null &&
               root.right.right == null))
            {
                System.out.print(root.data + " ");
            }
        }

        // If node has only left child
        if (root.left != null &&
           root.right == null)
        {
            if (root.left.left == null &&
               root.left.right == null)
            {
                System.out.print(root.data + " ");
            }
        }

        // If node has only right child
        if (root.right != null &&
             root.left == null)
        {
            if (root.right.left == null &&
               root.right.right == null)
            {
                System.out.print(root.data + " ");
            }
        }
    }

    // Recursively call for left
    // and right subtree
    cal(root.left);
    cal(root.right);
}

// Driver Code
public static void main (String[] args)
{
    GFG tree = new GFG();

    tree.root = new Node(20);
    tree.root.left = new Node(8);
    tree.root.right = new Node(22);

    tree.root.left.left = new Node(4);
    tree.root.left.right = new Node(12);

    tree.root.left.right.left = new Node(10);
    tree.root.left.right.right = new Node(14);

    // Function call
    cal(tree.root);
}
}

// This code is contributed by offbeat

Python3

# Python3 program for the
# above approach

# Node of tree
class newNode:

    def __init__(self, data):

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

# Creates and initializes a new
# node for the tree

# Prints all nodes which are
# just above leaf node
def cal(root):

    # If tree is empty
    if (root == None):
        return

    # If it is a leaf node
    if (root.left == None and
        root.right == None):
        return

    # For internal nodes
    else:

        # If node has two children
        if (root.left != None and
            root.right != None):
            if ((root.left.left == None and
                 root.left.right == None) or
                (root.right.left == None and
                 root.right.right == None)):
                print(root.data, end = " ")

        # If node has only left child
        if (root.left != None and
            root.right == None):
            if (root.left.left == None and
                root.left.right == None):
                print(root.data, end = " ")

        # If node has only right child
        if (root.right != None and
            root.left == None):
            if (root.right.left == None and
                root.right.right == None):
                print(root.data, end = " ")

    # Recursively Call for left
    # and right subtree
    cal(root.left)
    cal(root.right)

# Driver Code
if __name__ == '__main__':

    # Construct a tree
    root = newNode(20)
    root.left = newNode(8)
    root.right = newNode(22)
    root.left.left = newNode(4)
    root.left.right = newNode(12)
    root.left.right.left = newNode(10)
    root.left.right.right = newNode(14)

    # Function Call
    cal(root)

# This code is contributed by SURENDRA_GANGWAR

C#

// C# program for the above approach
using System;

// Class containing the left and right
// child of current node and the
// key value
class Node
{
    public int data;
    public Node left, right;

    // Constructor of the class
    public Node(int item)
    {
        data = item;
        left = right = null;
    }
}

class GFG{
Node root;

// Prints all nodes which are just
// above leaf node
static void cal(Node root)
{
    // If tree is empty
    if (root == null)
    {
        return;
    }

    // If it is a leaf node
    if (root.left == null &&
        root.right == null)
    {
        return;
    }

    // For internal nodes
    else
    {
        // If node has two children
        if (root.left != null &&
            root.right != null)
        {
            if ((root.left.left == null &&
                 root.left.right == null) ||
                (root.right.left == null &&
                 root.right.right == null))
            {
                Console.Write(root.data + " ");
            }
        }

        // If node has only left child
        if (root.left != null &&
            root.right == null)
        {
            if (root.left.left == null &&
                root.left.right == null)
            {
                Console.Write(root.data + " ");
            }
        }

        // If node has only right child
        if (root.right != null &&
            root.left == null)
        {
            if (root.right.left == null &&
                root.right.right == null)
            {
                Console.Write(root.data + " ");
            }
        }
    }

    // Recursively call for left
    // and right subtree
    cal(root.left);
    cal(root.right);
}

// Driver Code
public static void Main(String[] args)
{
    GFG tree = new GFG();
    tree.root = new Node(20);
    tree.root.left = new Node(8);
    tree.root.right = new Node(22);
    tree.root.left.left = new Node(4);
    tree.root.left.right = new Node(12);
    tree.root.left.right.left = new Node(10);
    tree.root.left.right.right = new Node(14);

    // Function call
    cal(tree.root);
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>
    // Javascript implementation of the above approach

    // Class containing the left and right
    // child of current node and the
    // key value
    class Node
    {
        constructor(item) {
           this.data = item;
           this.left = this.right = null;
        }
    }

    class GFG
    {

    }

    let root;

    // Prints all nodes which are just
    // above leaf node
    function cal(root)
    {

        // If tree is empty
        if (root == null)
        {
            return;
        }

        // If it is a leaf node
        if (root.left == null &&
           root.right == null)
        {
            return;
        }

        // For internal nodes
        else
        {

            // If node has two children
            if (root.left != null &&
               root.right != null)
            {
                if ((root.left.left == null &&
                    root.left.right == null) ||
                   (root.right.left == null &&
                   root.right.right == null))
                {
                    document.write(root.data + " ");
                }
            }

            // If node has only left child
            if (root.left != null &&
               root.right == null)
            {
                if (root.left.left == null &&
                   root.left.right == null)
                {
                    document.write(root.data + " ");
                }
            }

            // If node has only right child
            if (root.right != null &&
                 root.left == null)
            {
                if (root.right.left == null &&
                   root.right.right == null)
                {
                    document.write(root.data + " ");
                }
            }
        }

        // Recursively call for left
        // and right subtree
        cal(root.left);
        cal(root.right);
    }

    let tree = new GFG();

    tree.root = new Node(20);
    tree.root.left = new Node(8);
    tree.root.right = new Node(22);

    tree.root.left.left = new Node(4);
    tree.root.left.right = new Node(12);

    tree.root.left.right.left = new Node(10);
    tree.root.left.right.right = new Node(14);

    // Function call
    cal(tree.root);

</script>

Output:

20 8 12

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