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

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++

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// C++ program to print leaf nodes from
// left to right using one stack
#include <bits/stdc++.h>
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<Node*> 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;
}

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Java

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// 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<Node> s = new Stack<Node> (); 
  
    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

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C#

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// 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<Node> s = new Stack<Node> (); 
  
    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

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Output:

4 5 6 7


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Improved By : prerna saini, 29AjayKumar