Print left and right leaf nodes separately in Binary Tree

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++

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// 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;
}

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Java

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// 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

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C#

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// 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

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Output:

Left leaf nodes
3 
Right leaf nodes
4 2

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