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++
// C++ program to print leaf nodes from// left to right using one stack#include <bits/stdc++.h>using namespace std;// Structure of binary treestruct Node { Node* left; Node* right; int data;};// Function to create a new nodeNode* 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 stackvoid 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 Codeint 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<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 |
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 stackdef 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 Coderoot = 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<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 |
Javascript
<script> // JavaScript program to print leaf nodes from // left to{ right using one stack // Structure of binary tree class Node { constructor(key) { this.left = null; this.right = null; this.data = key; } } // Function to create a new node function newNode(key) { let node = new Node(key); return node; } // Function to Print all the leaf nodes // of Binary tree using one stack function printLeafLeftToRight(p) { // stack to store the nodes let s = []; 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.length == 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[s.length - 1].right == null) { p = s[s.length - 1]; s.pop(); // Print the leaf node if (p.left == null) document.write(p.data + " "); } while (p == s[s.length - 1].right) { p = s[s.length - 1]; s.pop(); if (s.length == 0) break; } // If stack is not empty then assign p as // the stack's top node's right child if (s.length != 0) p = s[s.length - 1].right; else p = null; } } } } let 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); </script> |
Output:
4 5 6 7
Time Complexity: O(N)
Auxiliary Space: O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
