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. 
Auxiliary Space: O(n)

Iterative Method:

Following is a simple stack-based iterative method to print the leaf nodes from left to right.

• Create an empty stack ‘st’ and push the root node to stack.
• Do the following while stack is not empty.
  • Pop an item from the stack
  • If the node is a leaf node then print it.
  • Else:
    • If the right node is not NULL
      • push the right node into the stack
    • If the left node is not NULL
      • push the left node into the stack

Below is the implementation of the above approach. 

C++

// C++ program to print leaf nodes from left to right
#include <bits/stdc++.h>
using namespace std;

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

// fun to creates and returns a new node
Node* newNode(int data)
{
    Node* temp = new Node();
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}

// fun to print leaf nodes from left to right
void printleafNodes(Node* root)
{
    // base case
    if (!root)
        return;

    // implement iterative preorder traversal and start
    // storing leaf nodes.
    stack<Node*> st;
    st.push(root);

    while (!st.empty()) {
        root = st.top();
        st.pop();

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

        if (root->right)
            st.push(root->right);
        if (root->left)
            st.push(root->left);
    }
}

// Driver program to test above functions
int main()
{

    // create a binary tree
    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);

    // prints leaf nodes of the given tree
    printleafNodes(root);

    return 0;
}
// This Method and code is contributed by Modem Upendra

Java

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

public class GFG {

    // structure of node of Binary Tree
    static class Node {
        int data;
        Node left, right;

        Node(int item)
        {
            data = item;
            left = right = null;
        }
    }

    // fun to print leaf nodes from left to right
    static void printleafNodes(Node root)
    {
        // base case
        if (root == null)
            return;

        // implement iterative preorder traversal and start
        // storing leaf nodes.
        Stack<Node> st = new Stack<>();
        st.push(root);

        while (!st.isEmpty()) {
            root = st.peek();
            st.pop();

            // if node is leafnode, print its data
            if (root.left == null && root.right == null)
                System.out.print(root.data + " ");

            if (root.right != null)
                st.push(root.right);
            if (root.left != null)
                st.push(root.left);
        }
    }

    // Driver program to test above functions
    public static void main(String[] args)
    {
        // create a binary tree
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.right.left = new Node(5);
        root.right.right = new Node(8);
        root.right.left.left = new Node(6);
        root.right.left.right = new Node(7);
        root.right.right.left = new Node(9);
        root.right.right.right = new Node(10);

        // prints leaf nodes of the given tree
        printleafNodes(root);
    }
}
// This code is contributed by Karandeep1234

Python3

# Python3 program to print leaf nodes from left to right

# A Binary Tree Node
class Node:

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

# fun to creates and returns a new node
def newNode(data):
    temp = Node(data)
    return temp

# fun to print leaf nodes from left to right
def printleafNodes(root):
    # base case
    if not root:
        return
    # implement iterative preorder traversal and start
    # storing leaf nodes.
    st = []
    st.append(root)

    while len(st) > 0:
        root = st.pop()

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

        if root.right:
            st.append(root.right)
        if root.left:
            st.append(root.left)

# Driver program to test above functions
if __name__ == '__main__':

    # create a binary tree
    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)

    # prints leaf nodes of the given tree
    printleafNodes(root)

# This code is contributed by lokeshpotta20.

C#

using System;
using System.Collections.Generic;

class GFG {
  // structure of node of Binary Tree
  class Node {
    public int data;
    public Node left, right;

    public Node(int item)
    {
      data = item;
      left = right = null;
    }
  }

  // fun to print leaf nodes from left to right
  static void printleafNodes(Node root)
  {
    // base case
    if (root == null)
      return;

    // implement iterative preorder traversal and start
    // storing leaf nodes.
    Stack<Node> st = new Stack<Node>();
    st.Push(root);

    while (st.Count != 0) {
      root = st.Peek();
      st.Pop();

      // if node is leafnode, print its data
      if (root.left == null && root.right == null)
        Console.Write(root.data + " ");

      if (root.right != null)
        st.Push(root.right);
      if (root.left != null)
        st.Push(root.left);
    }
  }

  // Driver program to test above functions
  public static void Main(string[] args)
  {
    // create a binary tree
    Node root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(3);
    root.left.left = new Node(4);
    root.right.left = new Node(5);
    root.right.right = new Node(8);
    root.right.left.left = new Node(6);
    root.right.left.right = new Node(7);
    root.right.right.left = new Node(9);
    root.right.right.right = new Node(10);

    // prints leaf nodes of the given tree
    printleafNodes(root);
  }
}

// This code is contributed by pradeepkumarppk2003

Javascript

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

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

// fun to print leaf nodes from left to right
function printleafNodes(root)
{
    // base case
    if (!root)
        return;

    // implement iterative preorder traversal and start
    // storing leaf nodes.
    let st=[];
    st.push(root);

    while (st.length>0) {
        root = st[st.length-1];
        st.pop();

        // if node is leafnode, print its data
        if (!root.left && !root.right)
            document.write(root.data + " ");

        if (root.right)
            st.push(root.right);
        if (root.left)
            st.push(root.left);
    }
}

// Driver program to test above functions
// create a binary tree
let root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.right.left = new Node(5);
root.right.right = new Node(8);
root.right.left.left = new Node(6);
root.right.left.right = new Node(7);
root.right.right.left = new Node(9);
root.right.right.right = new Node(10);

// prints leaf nodes of the given tree
printleafNodes(root);

Output

4 6 7 9 10 

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

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