Related Articles

# Print left and right leaf nodes separately in Binary Tree

• Difficulty Level : Medium
• Last Updated : 22 Jun, 2021

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 ``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 left_leaf,``            ``Vector 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 left_leaf,``                    ``Vector 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 left_leaf = ``new` `Vector(),``                ``right_leaf = ``new` `Vector();``        ``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

 ``
Output:
```Left leaf nodes
3
Right leaf nodes
4 2```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up