Print all leaf nodes of a Binary Tree from left to right

Given a binary tree, we need to write a program to print all leaf nodes of the given binary tree from left to right. That is, the nodes should be printed in the order they appear from left to right in the given tree. 

For Example, 

For the above binary tree, the output will be as shown below: 

4 6 7 9 10

The idea to do this is similar to DFS algorithm. Below is a step by step algorithm to do this: 

1. Check if the given node is null. If null, then return from the function.
2. Check if it is a leaf node. If the node is a leaf node, then print its data.
3. If in the above step, the node is not a leaf node then check if the left and right children of node exist. If yes then call the function for left and right child of the node recursively.

Below is the implementation of the above approach. 

C++

/* C++ program to print leaf nodes from left
   to right */
#include <iostream>
using namespace std;

// A Binary Tree Node
struct Node
{
    int data;
    struct Node *left, *right;
};

// function to print leaf
// nodes from left to right
void printLeafNodes(Node *root)
{
    // if node is null, return
    if (!root)
        return;

    // if node is leaf node, print its data
    if (!root->left && !root->right)
    {
        cout << root->data << " ";
        return;
    }

    // if left child exists, check for leaf
    // recursively
    if (root->left)
       printLeafNodes(root->left);

    // if right child exists, check for leaf
    // recursively
    if (root->right)
       printLeafNodes(root->right);
}

// Utility function to create a new tree node
Node* newNode(int data)
{
    Node *temp = new Node;
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}

// Driver program to test above functions
int main()
{
    // Let us create binary tree shown in
    // above diagram
    Node *root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->right->left = newNode(5);
    root->right->right = newNode(8);
    root->right->left->left = newNode(6);
    root->right->left->right = newNode(7);
    root->right->right->left = newNode(9);
    root->right->right->right = newNode(10);

    // print leaf nodes of the given tree
    printLeafNodes(root);

    return 0;
}

Java

// Java program to print leaf nodes
// from left to right
import java.util.*;

class GFG{

// A Binary Tree Node
static class Node
{
    public int data;
    public Node left, right;
};

// Function to print leaf
// nodes from left to right
static void printLeafNodes(Node root)
{

    // If node is null, return
    if (root == null)
        return;

    // If node is leaf node, print its data
    if (root.left == null &&
        root.right == null)
    {
        System.out.print(root.data + " ");
        return;
    }

    // If left child exists, check for leaf
    // recursively
    if (root.left != null)
        printLeafNodes(root.left);

    // If right child exists, check for leaf
    // recursively
    if (root.right != null)
        printLeafNodes(root.right);
}

// Utility function to create a new tree node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = null;
    temp.right = null;
    return temp;
}

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

    // Let us create binary tree shown in
    // above diagram
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.right.left = newNode(5);
    root.right.right = newNode(8);
    root.right.left.left = newNode(6);
    root.right.left.right = newNode(7);
    root.right.right.left = newNode(9);
    root.right.right.right = newNode(10);

    // Print leaf nodes of the given tree
    printLeafNodes(root);
}
}

// This code is contributed by pratham76

Python3

# Python3 program to print
# leaf nodes from left to right

# Binary tree node
class Node:

    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None

# Function to print leaf
# nodes from left to right
def printLeafNodes(root: Node) -> None:

    # If node is null, return
    if (not root):
        return

    # If node is leaf node,
    # print its data
    if (not root.left and
        not root.right):
        print(root.data,
              end = " ")
        return

    # If left child exists,
    # check for leaf recursively
    if root.left:
        printLeafNodes(root.left)

    # If right child exists,
    # check for leaf recursively
    if root.right:
        printLeafNodes(root.right)

# Driver Code
if __name__ == "__main__":

    # Let us create binary tree shown in
    # above diagram
    root = Node(1)
    root.left = Node(2)
    root.right = Node(3)
    root.left.left = Node(4)
    root.right.left = Node(5)
    root.right.right = Node(8)
    root.right.left.left = Node(6)
    root.right.left.right = Node(7)
    root.right.right.left = Node(9)
    root.right.right.right = Node(10)

    # print leaf nodes of the given tree
    printLeafNodes(root)

# This code is contributed by sanjeev2552

C#

// C# program to print leaf nodes
// from left to right
using System;

class GFG{

// A Binary Tree Node
class Node
{
    public int data;
    public Node left, right;
};

// Function to print leaf
// nodes from left to right
static void printLeafNodes(Node root)
{

    // If node is null, return
    if (root == null)
        return;

    // If node is leaf node, print its data
    if (root.left == null &&
        root.right == null)
    {
        Console.Write(root.data + " ");
        return;
    }

    // If left child exists, check for leaf
    // recursively
    if (root.left != null)
        printLeafNodes(root.left);

    // If right child exists, check for leaf
    // recursively
    if (root.right != null)
        printLeafNodes(root.right);
}

// Utility function to create a new tree node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = null;
    temp.right = null;
    return temp;
}

// Driver code
public static void Main()
{

    // Let us create binary tree shown in
    // above diagram
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.right.left = newNode(5);
    root.right.right = newNode(8);
    root.right.left.left = newNode(6);
    root.right.left.right = newNode(7);
    root.right.right.left = newNode(9);
    root.right.right.right = newNode(10);

    // Print leaf nodes of the given tree
    printLeafNodes(root);
}
}

// This code is contributed by rutvik_56

Javascript

<script>

// Javascript program to print leaf nodes
// from left to right

// A Binary Tree Node
class Node
{
    constructor()
    {
        this.data = 0;
        this.left = null;
        this.right = null;
    }
};

// Function to print leaf
// nodes from left to right
function printLeafNodes(root)
{

    // If node is null, return
    if (root == null)
        return;

    // If node is leaf node, print its data
    if (root.left == null &&
        root.right == null)
    {
        document.write(root.data + " ");
        return;
    }

    // If left child exists, check for leaf
    // recursively
    if (root.left != null)
        printLeafNodes(root.left);

    // If right child exists, check for leaf
    // recursively
    if (root.right != null)
        printLeafNodes(root.right);
}

// Utility function to create a new tree node
function newNode(data)
{
    var temp = new Node();
    temp.data = data;
    temp.left = null;
    temp.right = null;
    return temp;
}

// Driver code

// Let us create binary tree shown in
// above diagram
var root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.right.left = newNode(5);
root.right.right = newNode(8);
root.right.left.left = newNode(6);
root.right.left.right = newNode(7);
root.right.right.left = newNode(9);
root.right.right.right = newNode(10);

// Print leaf nodes of the given tree
printLeafNodes(root);

// This code is contributed by rrrtnx

</script>

Output:  

4 6 7 9 10

Time Complexity: O( n ) , where n is the number of nodes in the binary tree. 

https://youtu.be/rBekvLt23wY?list=PLqM7alHXFySHCXD7r1J0ky9Zg_GBB1dbk
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