# Print leaf nodes in binary tree from left to right using one stack

Given a binary tree, the task is 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.

Examples:

```Input :
1
/  \
2    3
/ \  / \
4  5  6  7
Output : 4 5 6 7

Input :
4
/  \
5    9
/ \  / \
8   3 7  2
/         / \
12        6   1
Output : 12 3 7 6 1
```

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

We have already discussed the iterative approach using two stacks.

Approach:The idea is to perform iterative postorder traversal using one stack and print the leaf nodes.

Below is the implementation of the above approach:

## C++

 `// C++ program to print leaf nodes from ` `// left to right using one stack ` `#include ` `using` `namespace` `std; ` ` `  `// Structure of binary tree ` `struct` `Node { ` `    ``Node* left; ` `    ``Node* right; ` `    ``int` `data; ` `}; ` ` `  `// Function to create a new node ` `Node* newNode(``int` `key) ` `{ ` `    ``Node* node = ``new` `Node(); ` `    ``node->left = node->right = NULL; ` `    ``node->data = key; ` `    ``return` `node; ` `} ` ` `  `// Function to Print all the leaf nodes ` `// of Binary tree using one stack ` `void` `printLeafLeftToRight(Node* p) ` `{ ` `    ``// stack to store the nodes ` `    ``stack s; ` ` `  `    ``while` `(1) { ` `        ``// If p is not null then push ` `        ``// it on the stack ` `        ``if` `(p) { ` `            ``s.push(p); ` `            ``p = p->left; ` `        ``} ` ` `  `        ``else` `{ ` `            ``// If stack is empty then come out ` `            ``// of the loop ` `            ``if` `(s.empty()) ` `                ``break``; ` `            ``else` `{ ` `                ``// If the node on top of the stack has its ` `                ``// right subtree as null then pop that node and ` `                ``// print the node only if its left ` `                ``// subtree is also null ` `                ``if` `(s.top()->right == NULL) { ` `                    ``p = s.top(); ` `                    ``s.pop(); ` ` `  `                    ``// Print the leaf node ` `                    ``if` `(p->left == NULL) ` `                        ``printf``(``"%d "``, p->data); ` `                ``} ` ` `  `                ``while` `(p == s.top()->right) { ` `                    ``p = s.top(); ` `                    ``s.pop(); ` ` `  `                    ``if` `(s.empty()) ` `                        ``break``; ` `                ``} ` ` `  `                ``// If stack is not empty then assign p as ` `                ``// the stack's top node's right child ` `                ``if` `(!s.empty()) ` `                    ``p = s.top()->right; ` `                ``else` `                    ``p = NULL; ` `            ``} ` `        ``} ` `    ``} ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``Node* root = newNode(1); ` `    ``root->left = newNode(2); ` `    ``root->right = newNode(3); ` `    ``root->left->left = newNode(4); ` `    ``root->left->right = newNode(5); ` `    ``root->right->left = newNode(6); ` `    ``root->right->right = newNode(7); ` ` `  `    ``printLeafLeftToRight(root); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to print leaf nodes from  ` `// left to{ right using one stack  ` `import` `java.util.*; ` `class` `GfG  ` `{  ` ` `  `// Structure of binary tree  ` `static` `class` `Node  ` `{  ` `    ``Node left;  ` `    ``Node right;  ` `    ``int` `data;  ` `} ` ` `  `// Function to create a new node  ` `static` `Node newNode(``int` `key)  ` `{  ` `    ``Node node = ``new` `Node();  ` `    ``node.left = ``null``; ` `    ``node.right = ``null``;  ` `    ``node.data = key;  ` `    ``return` `node;  ` `}  ` ` `  `// Function to Print all the leaf nodes  ` `// of Binary tree using one stack  ` `static` `void` `printLeafLeftToRight(Node p)  ` `{  ` `    ``// stack to store the nodes  ` `    ``Stack s = ``new` `Stack ();  ` ` `  `    ``while` `(``true``)  ` `    ``{  ` `        ``// If p is not null then push  ` `        ``// it on the stack  ` `        ``if` `(p != ``null``)  ` `        ``{  ` `            ``s.push(p);  ` `            ``p = p.left;  ` `        ``}  ` ` `  `        ``else` `        ``{  ` `            ``// If stack is empty then come out  ` `            ``// of the loop  ` `            ``if` `(s.isEmpty())  ` `                ``break``;  ` `            ``else`  `            ``{  ` `                ``// If the node on top of the stack has its  ` `                ``// right subtree as null then pop that node and  ` `                ``// print the node only if its left  ` `                ``// subtree is also null  ` `                ``if` `(s.peek().right == ``null``)  ` `                ``{  ` `                    ``p = s.peek();  ` `                    ``s.pop();  ` ` `  `                    ``// Print the leaf node  ` `                    ``if` `(p.left == ``null``)  ` `                        ``System.out.print(p.data + ``" "``);  ` `                ``}  ` ` `  `                ``while` `(p == s.peek().right)  ` `                ``{  ` `                    ``p = s.peek();  ` `                    ``s.pop();  ` ` `  `                    ``if` `(s.isEmpty())  ` `                        ``break``;  ` `                ``}  ` ` `  `                ``// If stack is not empty then assign p as  ` `                ``// the stack's top node's right child  ` `                ``if` `(!s.isEmpty())  ` `                    ``p = s.peek().right;  ` `                ``else` `                    ``p = ``null``;  ` `            ``}  ` `        ``}  ` `    ``}  ` `}  ` ` `  `// Driver Code  ` `public` `static` `void` `main(String[] args)  ` `{  ` `    ``Node root = newNode(``1``);  ` `    ``root.left = newNode(``2``);  ` `    ``root.right = newNode(``3``);  ` `    ``root.left.left = newNode(``4``);  ` `    ``root.left.right = newNode(``5``);  ` `    ``root.right.left = newNode(``6``);  ` `    ``root.right.right = newNode(``7``);  ` ` `  `    ``printLeafLeftToRight(root);  ` `}  ` `}  ` ` `  `// This code is contributed by Prerna Saini `

## Python3

 `# Python3 program to print leaf nodes from ` `# left to right using one stack ` ` `  `# Binary tree node  ` `class` `newNode:  ` `     `  `    ``def` `__init__(``self``, data):  ` `        ``self``.data ``=` `data  ` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None` ` `  `# Function to Print all the leaf nodes ` `# of Binary tree using one stack ` `def` `printLeafLeftToRight(p): ` `     `  `    ``# stack to store the nodes ` `    ``s ``=` `[] ` `    ``while` `(``1``): ` `         `  `        ``# If p is not None then push ` `        ``# it on the stack ` `        ``if` `(p): ` `            ``s.insert(``0``, p) ` `            ``p ``=` `p.left ` `         `  `        ``else``: ` `             `  `            ``# If stack is empty then come out ` `            ``# of the loop ` `            ``if` `len``(s) ``=``=` `0``: ` `                ``break` `             `  `            ``else``: ` `                 `  `                ``# If the node on top of the stack has its ` `                ``# right subtree as None then pop that node  ` `                ``# and print the node only if its left ` `                ``# subtree is also None ` `                ``if` `(s[``0``].right ``=``=` `None``): ` `                    ``p ``=` `s[``0``] ` `                    ``s.pop(``0``) ` `                     `  `                    ``# Print the leaf node ` `                    ``if` `(p.left ``=``=` `None``): ` `                        ``print``(p.data, end ``=` `" "``) ` `                 `  `                ``while` `(p ``=``=` `s[``0``].right): ` `                    ``p ``=` `s[``0``] ` `                    ``s.pop(``0``) ` `                     `  `                    ``if` `len``(s) ``=``=` `0``: ` `                        ``break` `                 `  `                ``# If stack is not empty then assign p as ` `                ``# the stack's top node's right child ` `                ``if` `len``(s): ` `                    ``p ``=` `s[``0``].right ` `                 `  `                ``else``: ` `                    ``p ``=` `None` ` `  `# Driver Code ` `root ``=` `newNode(``1``) ` `root.left ``=` `newNode(``2``) ` `root.right ``=` `newNode(``3``) ` `root.left.left ``=` `newNode(``4``) ` `root.left.right ``=` `newNode(``5``) ` `root.right.left ``=` `newNode(``6``) ` `root.right.right ``=` `newNode(``7``) ` ` `  `printLeafLeftToRight(root) ` ` `  `# This code is contributed by SHUBHAMSINGH10 `

## C#

 `// C# program to print leaf nodes from  ` `// left to{ right using one stack  ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GfG  ` `{  ` ` `  `// Structure of binary tree  ` `public` `class` `Node  ` `{  ` `    ``public` `Node left;  ` `    ``public` `Node right;  ` `    ``public` `int` `data;  ` `} ` ` `  `// Function to create a new node  ` `static` `Node newNode(``int` `key)  ` `{  ` `    ``Node node = ``new` `Node();  ` `    ``node.left = ``null``; ` `    ``node.right = ``null``;  ` `    ``node.data = key;  ` `    ``return` `node;  ` `}  ` ` `  `// Function to Print all the leaf nodes  ` `// of Binary tree using one stack  ` `static` `void` `printLeafLeftToRight(Node p)  ` `{  ` `    ``// stack to store the nodes  ` `    ``Stack s = ``new` `Stack ();  ` ` `  `    ``while` `(``true``)  ` `    ``{  ` `        ``// If p is not null then push  ` `        ``// it on the stack  ` `        ``if` `(p != ``null``)  ` `        ``{  ` `            ``s.Push(p);  ` `            ``p = p.left;  ` `        ``}  ` ` `  `        ``else` `        ``{  ` `            ``// If stack is empty then come out  ` `            ``// of the loop  ` `            ``if` `(s.Count == 0)  ` `                ``break``;  ` `            ``else` `            ``{  ` `                ``// If the node on top of the stack has its  ` `                ``// right subtree as null then pop that node and  ` `                ``// print the node only if its left  ` `                ``// subtree is also null  ` `                ``if` `(s.Peek().right == ``null``)  ` `                ``{  ` `                    ``p = s.Peek();  ` `                    ``s.Pop();  ` ` `  `                    ``// Print the leaf node  ` `                    ``if` `(p.left == ``null``)  ` `                        ``Console.Write(p.data + ``" "``);  ` `                ``}  ` ` `  `                ``while` `(p == s.Peek().right)  ` `                ``{  ` `                    ``p = s.Peek();  ` `                    ``s.Pop();  ` ` `  `                    ``if` `(s.Count == 0)  ` `                        ``break``;  ` `                ``}  ` ` `  `                ``// If stack is not empty then assign p as  ` `                ``// the stack's top node's right child  ` `                ``if` `(s.Count != 0)  ` `                    ``p = s.Peek().right;  ` `                ``else` `                    ``p = ``null``;  ` `            ``}  ` `        ``}  ` `    ``}  ` `}  ` ` `  `// Driver Code  ` `public` `static` `void` `Main(String[] args)  ` `{  ` `    ``Node root = newNode(1);  ` `    ``root.left = newNode(2);  ` `    ``root.right = newNode(3);  ` `    ``root.left.left = newNode(4);  ` `    ``root.left.right = newNode(5);  ` `    ``root.right.left = newNode(6);  ` `    ``root.right.right = newNode(7);  ` ` `  `    ``printLeafLeftToRight(root);  ` `}  ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

Output:

```4 5 6 7
```

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