# 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

```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++ 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 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 `

 `// 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 `

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

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Improved By : princiraj1992

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