Print left and right leaf nodes separately in Binary Tree

Namachila November 19, 2025

0 5 6 minutes read

Given a binary tree, the task is to print left and right leaf nodes separately.

Examples:

Input:
       0
     /   \
   1      2
 /  \
3    4 
Output:
Left Leaf Nodes: 3
Right Leaf Nodes: 4 2

Input:
   0
     \
      1
       \
        2
         \
          3
Output:
Left Leaf Nodes: None
Right Leaf Nodes: 3

Approach:

Check if given node is null. If null, then return from the function.
For each traversal at right and left, send information about the child (left or right child) using the parameter type. Set type = 0 while descending to the left branch and set type = 1 for the right branch.
Check if it is a leaf node. If the node is a leaf node, then store the leaf node in one of the two vectors of left and right child.
If node is not a leaf node continue traversal.
In the case of a single node tree, it will be both a root and a leaf node. This case has to be handled separately.

Below is the implementation of the above approach:

C++

// C++ program for the
// above approach
 
#include <bits/stdc++.h>
 
using namespace std;
 
// Structure for
// Binary Tree Node
struct Node {
    int data;
    Node* left;
    Node* right;
    Node(int x)
        : data(x)
        , left(NULL)
        , right(NULL)
    {
    }
};
 
// Function for
// dfs traversal
void dfs(Node* root, int type, vector<int>& left_leaf,
         vector<int>& right_leaf)
{
    // If node is
    // null, return
    if (!root) {
        return;
    }
 
    // If tree consists
    // of a single node
    if (!root->left && !root->right) {
        if (type == -1) {
            cout << "Tree consists of a single node\n";
        }
        else if (type == 0) {
            left_leaf.push_back(root->data);
        }
        else {
            right_leaf.push_back(root->data);
        }
 
        return;
    }
 
    // If left child exists,
    // traverse and set type to 0
    if (root->left) {
        dfs(root->left, 0, left_leaf, right_leaf);
    }
    // If right child exists,
    // traverse and set type to 1
    if (root->right) {
        dfs(root->right, 1, left_leaf, right_leaf);
    }
}
 
// Function to print
// the solution
void print(vector<int>& left_leaf, vector<int>& right_leaf)
{
 
    if (left_leaf.size() == 0 && right_leaf.size() == 0)
        return;
 
    // Printing left leaf nodes
    cout << "Left leaf nodes\n";
    for (int x : left_leaf) {
        cout << x << " ";
    }
    cout << '\n';
 
    // Printing right leaf nodes
    cout << "Right leaf nodes\n";
    for (int x : right_leaf) {
        cout << x << " ";
    }
    cout << '\n';
}
 
// Driver code
int main()
{
 
    Node* root = new Node(0);
    root->left = new Node(1);
    root->right = new Node(2);
    root->left->left = new Node(3);
    root->left->right = new Node(4);
 
    vector<int> left_leaf, right_leaf;
    dfs(root, -1, left_leaf, right_leaf);
 
    print(left_leaf, right_leaf);
 
    return 0;
}

Java

// Java program for the
// above approach
import java.util.*;
 
class GFG 
{
 
    // Structure for
    // Binary Tree Node
    static class Node 
    {
        int data;
        Node left;
        Node right;
 
        public Node(int data) 
        {
            this.data = data;
            this.left = null;
            this.right = null;
        }
 
    };
 
    // Function for
    // dfs traversal
    static void dfs(Node root, int type, Vector<Integer> left_leaf,
            Vector<Integer> right_leaf) 
    {
        // If node is
        // null, return
        if (root == null)
        {
            return;
        }
 
        // If tree consists
        // of a single node
        if (root.left == null && root.right == null)
        {
            if (type == -1) 
            {
                System.out.print("Tree consists of a single node\n");
            } 
            else if (type == 0)
            {
                left_leaf.add(root.data);
            }
            else
            {
                right_leaf.add(root.data);
            }
 
            return;
        }
 
        // If left child exists,
        // traverse and set type to 0
        if (root.left != null) 
        {
            dfs(root.left, 0, left_leaf, right_leaf);
        }
         
        // If right child exists,
        // traverse and set type to 1
        if (root.right != null) 
        {
            dfs(root.right, 1, left_leaf, right_leaf);
        }
    }
 
    // Function to print
    // the solution
    static void print(Vector<Integer> left_leaf, 
                    Vector<Integer> right_leaf) 
    {
 
        if (left_leaf.size() == 0 && right_leaf.size() == 0)
            return;
 
        // Printing left leaf nodes
        System.out.print("Left leaf nodes\n");
        for (int x : left_leaf)
        {
            System.out.print(x + " ");
        }
        System.out.println();
 
        // Printing right leaf nodes
        System.out.print("Right leaf nodes\n");
        for (int x : right_leaf)
        {
            System.out.print(x + " ");
        }
        System.out.println();
    }
 
    // Driver code
    public static void main(String[] args) 
    {
 
        Node root = new Node(0);
        root.left = new Node(1);
        root.right = new Node(2);
        root.left.left = new Node(3);
        root.left.right = new Node(4);
 
        Vector<Integer> left_leaf = new Vector<Integer>(),
                right_leaf = new Vector<Integer>();
        dfs(root, -1, left_leaf, right_leaf);
 
        print(left_leaf, right_leaf);
 
    }
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 program for the
# above approach
  
# Structure for
# Binary Tree Node
class Node:
     
    def __init__(self, data):
         
        self.data = data
        self.left = None
        self.right = None
  
# Function for 
# dfs traversal
def dfs(root, type_t, left_leaf, 
        right_leaf):
 
    # If node is
    # null, return
    if (not root):
        return
  
    # If tree consists
    # of a single node
    if (not root.left and not root.right):
        if (type_t == -1):
            print("Tree consists of a single node")
         
        elif (type_t == 0):
            left_leaf.append(root.data)
         
        else:
            right_leaf.append(root.data)
  
        return
  
    # If left child exists,
    # traverse and set type_t to 0
    if (root.left):
        dfs(root.left, 0, left_leaf, 
            right_leaf)
 
    # If right child exists,
    # traverse and set type_t to 1
    if (root.right):
        dfs(root.right, 1, left_leaf,
            right_leaf)
     
# Function to print
# the solution
def prints(left_leaf, right_leaf):
     
    if (len(left_leaf) == 0 and
        len(right_leaf) == 0):
        return
  
    # Printing left leaf nodes
    print("Left leaf nodes")
     
    for x in left_leaf:
        print(x, end = ' ')
     
    print()
  
    # Printing right leaf nodes
    print("Right leaf nodes")
     
    for x in right_leaf:
        print(x, end = ' ')
         
    print()
  
# Driver code
if __name__=='__main__':
  
    root = Node(0)
    root.left = Node(1)
    root.right = Node(2)
    root.left.left = Node(3)
    root.left.right = Node(4)
  
    left_leaf = []
    right_leaf = []
    dfs(root, -1, left_leaf, right_leaf)
  
    prints(left_leaf, right_leaf)
  
# This code is contributed by pratham76

C#

// C# program for the
// above approach
using System;
using System.Collections.Generic;
 
class GFG 
{
 
    // Structure for
    // Binary Tree Node
    public class Node 
    {
        public int data;
        public Node left;
        public Node right;
 
        public Node(int data) 
        {
            this.data = data;
            this.left = null;
            this.right = null;
        }
 
    };
 
    // Function for
    // dfs traversal
    static void dfs(Node root, int type, List<int> left_leaf,
            List<int> right_leaf) 
    {
        // If node is
        // null, return
        if (root == null)
        {
            return;
        }
 
        // If tree consists
        // of a single node
        if (root.left == null && root.right == null)
        {
            if (type == -1) 
            {
                Console.Write("Tree consists of a single node\n");
            } 
            else if (type == 0)
            {
                left_leaf.Add(root.data);
            }
            else
            {
                right_leaf.Add(root.data);
            }
 
            return;
        }
 
        // If left child exists,
        // traverse and set type to 0
        if (root.left != null) 
        {
            dfs(root.left, 0, left_leaf, right_leaf);
        }
         
        // If right child exists,
        // traverse and set type to 1
        if (root.right != null) 
        {
            dfs(root.right, 1, left_leaf, right_leaf);
        }
    }
 
    // Function to print
    // the solution
    static void print(List<int> left_leaf, 
                    List<int> right_leaf) 
    {
 
        if (left_leaf.Count == 0 && right_leaf.Count == 0)
            return;
 
        // Printing left leaf nodes
        Console.Write("Left leaf nodes\n");
        foreach (int x in left_leaf)
        {
            Console.Write(x + " ");
        }
        Console.WriteLine();
 
        // Printing right leaf nodes
        Console.Write("Right leaf nodes\n");
        foreach (int x in right_leaf)
        {
            Console.Write(x + " ");
        }
        Console.WriteLine();
    }
 
    // Driver code
    public static void Main(String[] args) 
    {
 
        Node root = new Node(0);
        root.left = new Node(1);
        root.right = new Node(2);
        root.left.left = new Node(3);
        root.left.right = new Node(4);
 
        List<int> left_leaf = new List<int>(),
                right_leaf = new List<int>();
        dfs(root, -1, left_leaf, right_leaf);
 
        print(left_leaf, right_leaf);
 
    }
}
 
// This code is contributed by PrinciRaj1992

Javascript

<script>
  
// JavaScript program for the
// above approach
// Structure for
// Binary Tree Node
class Node 
{
    constructor(data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
};
// Function for
// dfs traversal
function dfs(root, type, left_leaf, right_leaf) 
{
    // If node is
    // null, return
    if (root == null)
    {
        return;
    }
    // If tree consists
    // of a single node
    if (root.left == null && root.right == null)
    {
        if (type == -1) 
        {
            document.write("Tree consists of a single node<br>");
        } 
        else if (type == 0)
        {
            left_leaf.push(root.data);
        }
        else
        {
            right_leaf.push(root.data);
        }
        return;
    }
    // If left child exists,
    // traverse and set type to 0
    if (root.left != null) 
    {
        dfs(root.left, 0, left_leaf, right_leaf);
    }
     
    // If right child exists,
    // traverse and set type to 1
    if (root.right != null) 
    {
        dfs(root.right, 1, left_leaf, right_leaf);
    }
}
// Function to print
// the solution
function print(left_leaf, right_leaf) 
{
    if (left_leaf.Count == 0 && right_leaf.Count == 0)
        return;
    // Printing left leaf nodes
    document.write("Left leaf nodes<br>");
 
    for(var x of left_leaf)
    {
        document.write(x + " ");
    }
    document.write("<br>");
 
    // Printing right leaf nodes
    document.write("Right leaf nodes<br>");
    for(var x of right_leaf)
    {
        document.write(x + " ");
    }
    document.write("<br>");
}
 
// Driver code
var root = new Node(0);
root.left = new Node(1);
root.right = new Node(2);
root.left.left = new Node(3);
root.left.right = new Node(4);
var left_leaf = [];
var right_leaf = [];
dfs(root, -1, left_leaf, right_leaf);
print(left_leaf, right_leaf);
 
</script>

Output:

Left leaf nodes
3 
Right leaf nodes
4 2

Time complexity: O(h) where h is height of given Binary Tree

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