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Print All Leaf Nodes of a Binary Tree from left to right | Set-2 ( Iterative Approach )

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Given a Binary Tree, the task is to print the leaf nodes from left to right. The nodes must be printed in the order they appear from left to right.
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 Recursive approach. Here we will solve the problem using two stacks.
Approach:The idea is to use two stacks, one to store all the nodes of the tree and the other one to store all the leaf nodes. We will pop the top node of the first stack. If the node has a left child, we will push it on top of the first stack, if it has a right child then we will push it onto the top of the first stack, but if the node is a leaf node then we will push it onto the top of the second stack. We will do it for all the nodes until we have traversed the Binary tree completely.
Then we will start popping the second stack and print all its elements until the stack gets empty.
Below is the implementation of the above approach:
 

C++




// C++ program to print all the leaf nodes
// of a Binary tree from left to right
#include <bits/stdc++.h>
using namespace std;
 
// Binary tree node
struct Node {
    Node* left;
    Node* right;
    int data;
};
 
// Function to create a new
// Binary node
Node* newNode(int data)
{
    Node* temp = new Node;
 
    temp->data = data;
    temp->left = NULL;
    temp->right = NULL;
 
    return temp;
}
 
// Function to Print all the leaf nodes
// of Binary tree using two stacks
void PrintLeafLeftToRight(Node* root)
{
    // Stack to store all the nodes of tree
    stack<Node*> s1;
 
    // Stack to store all the leaf nodes
    stack<Node*> s2;
 
    // Push the root node
    s1.push(root);
 
    while (!s1.empty()) {
        Node* curr = s1.top();
        s1.pop();
 
        // If current node has a left child
        // push it onto the first stack
        if (curr->left)
            s1.push(curr->left);
 
        // If current node has a right child
        // push it onto the first stack
        if (curr->right)
            s1.push(curr->right);
 
        // If current node is a leaf node
        // push it onto the second stack
        else if (!curr->left && !curr->right)
            s2.push(curr);
    }
 
    // Print all the leaf nodes
    while (!s2.empty()) {
        cout << s2.top()->data << " ";
        s2.pop();
    }
}
 
// Driver code
int main()
{
    Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->right->left = newNode(5);
    root->right->right = newNode(7);
    root->left->left->left = newNode(10);
    root->left->left->right = newNode(11);
    root->right->right->left = newNode(8);
 
    PrintLeafLeftToRight(root);
 
    return 0;
}


Java




// Java program to print all the leaf nodes
// of a Binary tree from left to right
import java.util.*;
 
class GFG
{
 
// Binary tree node
static class Node
{
    Node left;
    Node right;
    int data;
};
 
// Function to create a new
// Binary node
static Node newNode(int data)
{
    Node temp = new Node();
 
    temp.data = data;
    temp.left = null;
    temp.right = null;
 
    return temp;
}
 
// Function to Print all the leaf nodes
// of Binary tree using two stacks
static void PrintLeafLeftToRight(Node root)
{
    // Stack to store all the nodes of tree
    Stack<Node> s1 = new Stack<>();
 
    // Stack to store all the leaf nodes
    Stack<Node> s2 = new Stack<>();;
 
    // Push the root node
    s1.push(root);
 
    while (!s1.empty())
    {
        Node curr = s1.peek();
        s1.pop();
 
        // If current node has a left child
        // push it onto the first stack
        if (curr.left!=null)
            s1.push(curr.left);
 
        // If current node has a right child
        // push it onto the first stack
        if (curr.right!=null)
            s1.push(curr.right);
 
        // If current node is a leaf node
        // push it onto the second stack
        else if (curr.left==null && curr.right==null)
            s2.push(curr);
    }
 
    // Print all the leaf nodes
    while (!s2.empty())
    {
        System.out.print(s2.peek().data +" ");
        s2.pop();
    }
}
 
// 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.right.left = newNode(5);
    root.right.right = newNode(7);
    root.left.left.left = newNode(10);
    root.left.left.right = newNode(11);
    root.right.right.left = newNode(8);
 
    PrintLeafLeftToRight(root);
}
}
 
// This code is contributed by 29AjayKumar


Python3




# Python3 program to print all the leaf
# nodes of a Binary tree from left to right
 
# Binary tree node
class Node:
     
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
 
# Function to Print all the leaf nodes
# of Binary tree using two stacks
def PrintLeafLeftToRight(root):
 
    # Stack to store all the nodes
    # of tree
    s1 = []
 
    # Stack to store all the
    # leaf nodes
    s2 = []
 
    # Push the root node
    s1.append(root)
 
    while len(s1) != 0:
        curr = s1.pop()
 
        # If current node has a left child
        # push it onto the first stack
        if curr.left:
            s1.append(curr.left)
 
        # If current node has a right child
        # push it onto the first stack
        if curr.right:
            s1.append(curr.right)
 
        # If current node is a leaf node
        # push it onto the second stack
        elif not curr.left and not curr.right:
            s2.append(curr)
     
    # Print all the leaf nodes
    while len(s2) != 0:
        print(s2.pop().data, end = " ")
     
# Driver code
if __name__ == "__main__":
 
    root = Node(1)
    root.left = Node(2)
    root.right = Node(3)
    root.left.left = Node(4)
    root.right.left = Node(5)
    root.right.right = Node(7)
    root.left.left.left = Node(10)
    root.left.left.right = Node(11)
    root.right.right.left = Node(8)
 
    PrintLeafLeftToRight(root)
 
# This code is contributed
# by Rituraj Jain


C#




// C# program to print all the leaf nodes
// of a Binary tree from left to right
using System;
using System.Collections.Generic;
 
class GFG
{
 
// Binary tree node
public class Node
{
    public Node left;
    public Node right;
    public int data;
};
 
// Function to create a new
// Binary node
static Node newNode(int data)
{
    Node temp = new Node();
 
    temp.data = data;
    temp.left = null;
    temp.right = null;
 
    return temp;
}
 
// Function to Print all the leaf nodes
// of Binary tree using two stacks
static void PrintLeafLeftToRight(Node root)
{
    // Stack to store all the nodes of tree
    Stack<Node> s1 = new Stack<Node>();
 
    // Stack to store all the leaf nodes
    Stack<Node> s2 = new Stack<Node>();;
 
    // Push the root node
    s1.Push(root);
 
    while (s1.Count != 0)
    {
        Node curr = s1.Peek();
        s1.Pop();
 
        // If current node has a left child
        // push it onto the first stack
        if (curr.left != null)
            s1.Push(curr.left);
 
        // If current node has a right child
        // push it onto the first stack
        if (curr.right != null)
            s1.Push(curr.right);
 
        // If current node is a leaf node
        // push it onto the second stack
        else if (curr.left == null && curr.right == null)
            s2.Push(curr);
    }
 
    // Print all the leaf nodes
    while (s2.Count != 0)
    {
        Console.Write(s2.Peek().data + " ");
        s2.Pop();
    }
}
 
// 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.right.left = newNode(5);
    root.right.right = newNode(7);
    root.left.left.left = newNode(10);
    root.left.left.right = newNode(11);
    root.right.right.left = newNode(8);
 
    PrintLeafLeftToRight(root);
}
}
 
// This code is contributed by Rajput-Ji


Javascript




<script>
 
    // JavaScript implementation of the approach
     
    // Binary tree node
    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }
     
    // Function to create a new
    // Binary node
    function newNode(data)
    {
        let temp = new Node(data);
        return temp;
    }
 
    // Function to Print all the leaf nodes
    // of Binary tree using two stacks
    function PrintLeafLeftToRight(root)
    {
        // Stack to store all the nodes of tree
        let s1 = [];
 
        // Stack to store all the leaf nodes
        let s2 = [];
 
        // Push the root node
        s1.push(root);
 
        while (s1.length > 0)
        {
            let curr = s1[s1.length - 1];
            s1.pop();
 
            // If current node has a left child
            // push it onto the first stack
            if (curr.left!=null)
                s1.push(curr.left);
 
            // If current node has a right child
            // push it onto the first stack
            if (curr.right!=null)
                s1.push(curr.right);
 
            // If current node is a leaf node
            // push it onto the second stack
            else if (curr.left==null && curr.right==null)
                s2.push(curr);
        }
 
        // Print all the leaf nodes
        while (s2.length > 0)
        {
            document.write(s2[s2.length - 1].data +" ");
            s2.pop();
        }
    }
     
    let root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.right.left = newNode(5);
    root.right.right = newNode(7);
    root.left.left.left = newNode(10);
    root.left.left.right = newNode(11);
    root.right.right.left = newNode(8);
   
    PrintLeafLeftToRight(root);
     
</script>


Output: 

10 11 5 8

 

Time Complexity: O(N) 
Auxiliary Space: O(N)



Last Updated : 06 Aug, 2021
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