Print nodes of a Binary Search Tree in Top Level Order and Reversed Bottom Level Order alternately

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Given a Binary Search Tree, the task is to print the nodes of the BST in the following order:

If the BST contains levels numbered from 1 to N then, the printing order is level 1, level N, level 2, level N – 1, and so on.
The top-level order (1, 2, …) nodes are printed from left to right, while the bottom level order (N, N-1, …) nodes are printed from right to left.

Examples:

Input:
Output: 27 42 31 19 10 14 35
Explanation:
Level 1 from left to right: 27
Level 3 from right to left: 42 31 19 10
Level 2 from left to right: 14 35

Input:
Output: 25 48 38 28 12 5 20 36 40 30 22 10

Approach: To solve the problem, the idea is to store the nodes of BST in ascending and descending order of levels and node values and print all the nodes of the same level alternatively between ascending and descending order. Follow the steps below to solve the problem:

Initialize a Min Heap and a Max Heap to store the nodes in ascending and descending order of level and node values respectively.
Perform a level order traversal on the given BST to store the nodes in the respective priority queues.
Print all the nodes of each level one by one from the Min Heap followed by the Max Heap alternately.
If any level in the Min Heap or Max Heap is found to be already printed, skip to the next level.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Structure of a BST node
struct node {
    int data;
    struct node* left;
    struct node* right;
};
 
// Utility function to create a new BST node
struct node* newnode(int d)
{
    struct node* temp
        = (struct node*)malloc(sizeof(struct node));
    temp->left = NULL;
    temp->right = NULL;
    temp->data = d;
    return temp;
}
 
// Function to print the nodes of a
// BST in Top Level Order and Reversed
// Bottom Level Order alternatively
void printBST(node* root)
{
    // Stores the nodes in descending order
    // of the level and node values
    priority_queue<pair<int, int> > great;
 
    // Stores the nodes in ascending order
    // of the level and node values
 
    priority_queue<pair<int, int>,
                   vector<pair<int, int> >,
                   greater<pair<int, int> > >
        small;
 
    // Initialize a stack for
    // level order traversal
    stack<pair<node*, int> > st;
 
    // Push the root of BST
    // into the stack
    st.push({ root, 1 });
 
    // Perform Level Order Traversal
    while (!st.empty()) {
 
        // Extract and pop the node
        // from the current level
        node* curr = st.top().first;
 
        // Stores level of current node
        int level = st.top().second;
        st.pop();
 
        // Store in the priority queues
        great.push({ level, curr->data });
        small.push({ level, curr->data });
 
        // Traverse left subtree
        if (curr->left)
            st.push({ curr->left, level + 1 });
 
        // Traverse right subtree
        if (curr->right)
            st.push({ curr->right, level + 1 });
    }
 
    // Stores the levels that are printed
    unordered_set<int> levelsprinted;
 
    // Print the nodes in the required manner
    while (!small.empty() && !great.empty()) {
 
        // Store the top level of traversal
        int toplevel = small.top().first;
 
        // If the level is already printed
        if (levelsprinted.find(toplevel)
            != levelsprinted.end())
            break;
 
        // Otherwise
        else
            levelsprinted.insert(toplevel);
 
        // Print nodes of same level
        while (!small.empty()
               && small.top().first == toplevel) {
            cout << small.top().second << " ";
            small.pop();
        }
 
        // Store the bottom level of traversal
        int bottomlevel = great.top().first;
 
        // If the level is already printed
        if (levelsprinted.find(bottomlevel)
            != levelsprinted.end()) {
            break;
        }
        else {
            levelsprinted.insert(bottomlevel);
        }
 
        // Print the nodes of same level
        while (!great.empty()
               && great.top().first == bottomlevel) {
            cout << great.top().second << " ";
            great.pop();
        }
    }
}
 
// Driver Code
int main()
{
    /*
    Given BST
 
                                25
                              /     \
                            20      36
                           /  \      / \
                          10   22   30 40
                         /  \      /   / \
                        5   12    28  38 48
    */
 
    // Creating the BST
    node* root = newnode(25);
    root->left = newnode(20);
    root->right = newnode(36);
    root->left->left = newnode(10);
    root->left->right = newnode(22);
    root->left->left->left = newnode(5);
    root->left->left->right = newnode(12);
    root->right->left = newnode(30);
    root->right->right = newnode(40);
    root->right->left->left = newnode(28);
    root->right->right->left = newnode(38);
    root->right->right->right = newnode(48);
 
    // Function Call
    printBST(root);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
public class GFG {
 
  // Structure of a BST node
  static class node {
    int data;
    node left;
    node right;
  }
 
  //Structure of pair (used in PriorityQueue)
  static class pair{
    int x,y;
    pair(int xx, int yy){
      this.x=xx;
      this.y=yy;
    }
  }
 
  //Structure of pair (used in Stack)
  static class StackPair{
    node n;
    int x;
    StackPair(node nn, int xx){
      this.n=nn;
      this.x=xx;
    }
  }
 
  // Utility function to create a new BST node
  static node newnode(int d)
  {
    node temp = new node();
    temp.left = null;
    temp.right = null;
    temp.data = d;
    return temp;
  }
 
  //Custom Comparator for pair class to sort
  //elements in increasing order 
  static class IncreasingOrder implements Comparator<pair>{
    public int compare(pair p1, pair p2){
      if(p1.x>p2.x){
        return 1;
      }else{
        if(p1.x<p2.x){
          return -1;
        }else{
          if(p1.y>p2.y){
            return 1;
          }else{
            if(p1.y<p2.y){
              return -1;
            }else{
              return 0;
            }
          }
        }
      }
    }
  }
 
  // Custom Comparator for pair class to sort
  // elements in decreasing order 
  static class DecreasingOrder implements Comparator<pair>{
    public int compare(pair p1, pair p2){
      if(p1.x>p2.x){
        return -1;
      }else{
        if(p1.x<p2.x){
          return 1;
        }else{
          if(p1.y>p2.y){
            return -1;
          }else{
            if(p1.y<p2.y){
              return 1;
            }else{
              return 0;
            }
          }
        }
      }
    }
  }
 
  // Function to print the nodes of a
  // BST in Top Level Order and Reversed
  // Bottom Level Order alternatively
  static void printBST(node root)
  {
    // Stores the nodes in descending order
    // of the level and node values
    PriorityQueue<pair> great = new PriorityQueue<>(new DecreasingOrder());
 
    // Stores the nodes in ascending order
    // of the level and node values
 
    PriorityQueue<pair> small = new PriorityQueue<>(new IncreasingOrder());
 
    // Initialize a stack for
    // level order traversal
    Stack<StackPair> st = new Stack<>();
 
    // Push the root of BST
    // into the stack
    st.push(new StackPair(root,1));
 
    // Perform Level Order Traversal
    while (!st.isEmpty()) {
 
      // Extract and pop the node
      // from the current level
      StackPair sp = st.pop();
      node curr = sp.n;
 
      // Stores level of current node
      int level = sp.x;
 
      // Store in the priority queues
      great.add(new pair(level,curr.data));
      small.add(new pair(level,curr.data));
 
      // Traverse left subtree
      if (curr.left!=null)
        st.push(new StackPair(curr.left,level+1));
 
      // Traverse right subtree
      if (curr.right!=null)
        st.push(new StackPair(curr.right,level+1));
    }
 
    // Stores the levels that are printed
    HashSet<Integer> levelsprinted = new HashSet<>();
 
    // Print the nodes in the required manner
    while (!small.isEmpty() && !great.isEmpty()) {
 
      // Store the top level of traversal
      int toplevel = small.peek().x;
 
      // If the level is already printed
      if (levelsprinted.contains(toplevel))
        break;
 
      // Otherwise
      else
        levelsprinted.add(toplevel);
 
      // Print nodes of same level
      while (!small.isEmpty() && small.peek().x == toplevel) {
        System.out.print(small.poll().y + " ");
      }
 
      // Store the bottom level of traversal
      int bottomlevel = great.peek().x;
 
      // If the level is already printed
      if (levelsprinted.contains(bottomlevel)) {
        break;
      }
      else {
        levelsprinted.add(bottomlevel);
      }
 
      // Print the nodes of same level
      while (!great.isEmpty() && great.peek().x == bottomlevel) {
        System.out.print(great.poll().y + " ");
      }
    }
  }
 
  public static void main (String[] args) {
    /*
        Given BST
 
                                    25
                                  /     \
                                20      36
                               /  \      / \
                              10   22   30 40
                             /  \      /   / \
                            5   12    28  38 48
        */
 
    // Creating the BST
    node root = newnode(25);
    root.left = newnode(20);
    root.right = newnode(36);
    root.left.left = newnode(10);
    root.left.right = newnode(22);
    root.left.left.left = newnode(5);
    root.left.left.right = newnode(12);
    root.right.left = newnode(30);
    root.right.right = newnode(40);
    root.right.left.left = newnode(28);
    root.right.right.left = newnode(38);
    root.right.right.right = newnode(48);
 
    // Function Call
    printBST(root);
  }
}
 
// This code is contributed by shruti456rawal

Python3

import queue
 
# Structure of a BST node
 
 
class Node:
    def __init__(self, d):
        self.data = d
        self.left = None
        self.right = None
 
# Function to print the nodes of a
# BST in Top Level Order and Reversed
# Bottom Level Order alternatively
 
 
def printBST(root):
    # Stores the nodes in descending order
    # of the level and node values
    great = queue.PriorityQueue()
 
    # Stores the nodes in ascending order
    # of the level and node values
    small = queue.PriorityQueue()
 
    # Initialize a queue for
    # level order traversal
    q = queue.Queue()
 
    # Push the root of BST
    # into the queue
    q.put((root, 1))
 
    # Perform Level Order Traversal
    while not q.empty():
 
        # Extract and pop the node
        # from the current level
        curr, level = q.get()
 
        # Store in the priority queues
        great.put((-level, curr.data))
        small.put((level, curr.data))
 
        # Traverse left subtree
        if curr.left:
            q.put((curr.left, level + 1))
 
        # Traverse right subtree
        if curr.right:
            q.put((curr.right, level + 1))
 
    # Stores the levels that are printed
    levelsprinted = set()
 
    # Print the nodes in the required manner
    while not small.empty() and not great.empty():
 
        # Store the top level of traversal
        toplevel = small.queue[0][0]
 
        # If the level is already printed
        if toplevel in levelsprinted:
            break
 
        # Otherwise
        else:
            levelsprinted.add(toplevel)
 
        # Print nodes of same level
        while not small.empty() and small.queue[0][0] == toplevel:
            print(small.get()[1], end=' ')
 
        # Store the bottom level of traversal
        bottomlevel = -great.queue[0][0]
 
        # If the level is already printed
        if bottomlevel in levelsprinted:
            break
        else:
            levelsprinted.add(bottomlevel)
 
        # Print the nodes of same level
        while not great.empty() and -great.queue[0][0] == bottomlevel:
            print(great.get()[1], end=' ')
 
 
# Driver Code
if __name__ == '__main__':
    """
    Given BST
 
                                25
                              /     \
                            20      36
                           /  \      / \
                          10   22   30 40
                         /  \      /   / \
                        5   12    28  38 48
    """
 
    # Creating the BST
    root = Node(25)
    root.left = Node(20)
    root.right = Node(36)
    root.left.left = Node(10)
    root.left.right = Node(22)
    root.left.left.left = Node(5)
    root.left.left.right = Node(12)
    root.right.left = Node(30)
    root.right.right = Node(40)
    root.right.left.left = Node(28)
    root.right.right.left = Node(38)
    root.right.right.right = Node(48)
 
    # Function Call
    printBST(root)

C#

using System;
using System.Collections.Generic;
 
class TreeNode {
    public int value;
    public TreeNode left, right;
    public TreeNode(int value)
    {
        this.value = value;
        left = null;
        right = null;
    }
}
 
class Program {
    static List<List<int> >
    GetLevelOrderTraversal(TreeNode root)
    {
        List<List<int> > ans = new List<List<int> >();
        // This will store values of nodes for the level
        // which we are currently traversing
        List<int> currentLevel = new List<int>();
 
        // We will be pushing null at the end of each level,
        // So whenever we encounter a null, it means we have
        // traversed all the nodes of the previous level
        Queue<TreeNode> q = new Queue<TreeNode>();
        q.Enqueue(root);
        q.Enqueue(null);
 
        while (q.Count > 1) {
            TreeNode currentNode = q.Dequeue();
            if (currentNode == null) {
                ans.Add(currentLevel);
                currentLevel = new List<int>();
                if (q.Count == 0) {
                    // It means no more level to be
                    // traversed
                    return ans;
                }
                else {
                    q.Enqueue(null);
                }
            }
            else {
                currentLevel.Add(currentNode.value);
                if (currentNode.left != null) {
                    q.Enqueue(currentNode.left);
                }
                if (currentNode.right != null) {
                    q.Enqueue(currentNode.right);
                }
            }
        }
        ans.Add(currentLevel);
        return ans;
    }
 
    static void Main(string[] args)
    {
        /*
         Given BST
 
                                     25
                                   /     \
                                 20      36
                                /  \      / \
                               10   22   30 40
                              /  \      /   / \
                             5   12    28  38 48
         */
 
        // Creating the BST
        TreeNode root = new TreeNode(25);
        root.left = new TreeNode(20);
        root.right = new TreeNode(36);
 
        root.left.left = new TreeNode(10);
        root.left.right = new TreeNode(22);
 
        root.left.left.left = new TreeNode(5);
        root.left.left.right = new TreeNode(12);
 
        root.right.left = new TreeNode(30);
        root.right.right = new TreeNode(40);
 
        root.right.left.left = new TreeNode(28);
        root.right.right.left = new TreeNode(38);
        root.right.right.right = new TreeNode(48);
 
        // Getting the value of nodes level wise
        List<List<int> > levelOrderTraversal
            = GetLevelOrderTraversal(root);
 
        // Now traversing the tree alternatively from top
        // and bottom using 2 pointers
        int i = 0;
        int j = levelOrderTraversal.Count - 1;
        while (i <= j) {
            if (i != j) {
                for (int k = 0;
                     k < levelOrderTraversal[i].Count;
                     k++) {
                    Console.Write(levelOrderTraversal[i][k]
                                  + " ");
                }
                for (int k
                     = levelOrderTraversal[j].Count - 1;
                     k >= 0; k--) {
                    Console.Write(levelOrderTraversal[j][k]
                                  + " ");
                }
            }
            else {
                // This will take care of the case when we
                // have odd number of levels in a BST
                for (int k = 0;
                     k < levelOrderTraversal[i].Count;
                     k++) {
                    Console.Write(levelOrderTraversal[i][k]
                                  + " ");
                }
            }
            i++;
            j--;
        }
    }
}
 
// This Code is contributed by Gaurav_Arora

Output:

25 48 38 28 12 5 20 36 40 30 22 10

Time Complexity: O(V log(V)), where V denotes the number of vertices in the given Binary Tree
Auxiliary Space: O(V)

Iterative Method(using queue):
Follow the steps to solve the given problem:
1). Perform level order traversal and keep track to level at each vertex of given tree.
2). Declare a queue to perform level order traversal and a vector of vector to store the level order traversal respectively to levels of given binary tree.
3). After storing all the vertex level wise we will initialize two iterator first will print data in level order and second will print the reverse level order and after printing the we will first first iterator and decrease the second iterator.

Below is the implementation of above approach:

C++

// C++ Program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// structure of tree node
struct Node {
    int data;
    struct Node* left;
    struct Node* right;
    Node(int data)
    {
        this->data = data;
        this->left = NULL;
        this->right = NULL;
    }
};
 
// function to find the height of binary tree
int height(Node* root)
{
    if (root == NULL)
        return 0;
    return max(height(root->left), height(root->right)) + 1;
}
 
// Function to print the nodes of a
// BST in Top Level Order and Reversed
// Bottom Level Order alternatively
void printBST(Node* root)
{
    // base cas
    if (root == NULL)
        return;
    vector<vector<int> > ans(height(root));
    int level = 0;
    // initializing the queue for level order traversal
    queue<Node*> q;
    q.push(root);
    while (!q.empty()) {
        int n = q.size();
        for (int i = 0; i < n; i++) {
            Node* front_node = q.front();
            q.pop();
            ans[level].push_back(front_node->data);
            if (front_node->left != NULL)
                q.push(front_node->left);
            if (front_node->right != NULL)
                q.push(front_node->right);
        }
        level++;
    }
    auto it1 = ans.begin();
    auto it2 = ans.end() - 1;
    while (it1 < it2) {
        for (int i : *it1) {
            cout << i << " ";
        }
        for (int i = (*it2).size() - 1; i >= 0; i--) {
            cout << (*it2)[i] << " ";
        }
        it1++;
        it2--;
    }
    if (it1 == it2) {
        for (int i : *it1) {
            cout << i << " ";
        }
    }
}
 
// driver code to test above function
int main()
{
    // creating the binary tree
    Node* root = new Node(25);
    root->left = new Node(20);
    root->right = new Node(36);
    root->left->left = new Node(10);
    root->left->right = new Node(22);
    root->left->left->left = new Node(5);
    root->left->left->right = new Node(12);
    root->right->left = new Node(30);
    root->right->right = new Node(40);
    root->right->left->left = new Node(28);
    root->right->right->left = new Node(38);
    root->right->right->right = new Node(48);
 
    printBST(root);
    return 0;
}
 
// THIS CODE IS CONTRIBUTED BY KIRTI
// AGARWAL(KIRTIAGARWAL23121999)

Java

import java.util.*;
 
// structure of tree node
class Node {
    int data;
    Node left, right;
    Node(int data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
 
// class to print the nodes of a
// BST in Top Level Order and Reversed
// Bottom Level Order alternatively
class Main {
    // function to find the height of binary tree
    public static int height(Node root)
    {
        if (root == null)
            return 0;
        return Math.max(height(root.left),
                        height(root.right))
            + 1;
    }
 
    public static void printBST(Node root)
    {
        // base case
        if (root == null)
            return;
 
        List<List<Integer> > ans = new ArrayList<>();
        int level = 0;
 
        // initializing the queue for level order traversal
        Queue<Node> q = new LinkedList<>();
        q.add(root);
 
        while (!q.isEmpty()) {
            int n = q.size();
            List<Integer> currLevel = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                Node front_node = q.poll();
                currLevel.add(front_node.data);
                if (front_node.left != null)
                    q.add(front_node.left);
                if (front_node.right != null)
                    q.add(front_node.right);
            }
            ans.add(currLevel);
            level++;
        }
 
        int it1 = 0;
        int it2 = ans.size() - 1;
        while (it1 < it2) {
            for (int i : ans.get(it1)) {
                System.out.print(i + " ");
            }
            for (int i = ans.get(it2).size() - 1; i >= 0;
                 i--) {
                System.out.print(ans.get(it2).get(i) + " ");
            }
            it1++;
            it2--;
        }
        if (it1 == it2) {
            for (int i : ans.get(it1)) {
                System.out.print(i + " ");
            }
        }
    }
 
    // driver code to test above function
    public static void main(String[] args)
    {
        // creating the binary tree
        Node root = new Node(25);
        root.left = new Node(20);
        root.right = new Node(36);
        root.left.left = new Node(10);
        root.left.right = new Node(22);
        root.left.left.left = new Node(5);
        root.left.left.right = new Node(12);
        root.right.left = new Node(30);
        root.right.right = new Node(40);
        root.right.left.left = new Node(28);
        root.right.right.left = new Node(38);
        root.right.right.right = new Node(48);
 
        printBST(root);
    }
}

Javascript

// JavaScript  program for the above approach
// structure of tree node
 
class Node{
    constructor(data){
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
 
// function to find the height of binary tree
function height(root){
    if(root == null) return 0;
    return Math.max(height(root.left), height(root.right)) + 1;
}
 
// function to print the nodes of a
// BST in Top level order and reversed
// bottom level order alternatively
function printBST(root){
    // base case
    if(root == null) return;
    let ans = [];
    for(let i = 0; i<height(root); i++){
        ans[i] = [];
    }
     
    let level = 0;
    // initializing the queue for level roder traversal
    let q = [];
    q.push(root);
    while(q.length > 0){
        let n = q.length;
        for(let i = 0; i<n; i++){
            let front_node = q.shift();
            ans[level].push(front_node.data);
            if(front_node.left) q.push(front_node.left);
            if(front_node.right) q.push(front_node.right);
        }
        level++;
    }
    let it1 = 0;
    let it2 = ans.length - 1;
    while(it1 < it2){
        for(let i = 0; i<ans[it1].length; i++){
            console.log(ans[it1][i] + " ");
        }
        for(let i = ans[it2].length -1; i >= 0; i--){
            console.log(ans[it2][i] + " ");
        }
        it1++;
        it2--;
    }
    if(it1 == it2){
        for(let i = 0; i<ans[it1].length; i++){
            console.log(ans[it1][i] + " ");
        }
    }
}
 
// driver program to test above function
let root = new Node(25);
root.left = new Node(20);
root.right = new Node(36);
root.left.left = new Node(10);
root.left.right = new Node(22);
root.left.left.left = new Node(5);
root.left.left.right = new Node(12);
root.right.left = new Node(30);
root.right.right = new Node(40);
root.right.left.left = new Node(28);
root.right.right.left = new Node(38);
root.right.right.right = new Node(48);
 
printBST(root);
 
// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)

Python

# Define the structure of tree node
class Node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
 
# Function to find the height of binary tree
 
 
def height(root):
    if root is None:
        return 0
    return max(height(root.left), height(root.right)) + 1
 
# Function to print the nodes of a BST in top-level order and reversed bottom-level order alternatively
 
 
def printBST(root):
    # Base case
    if root is None:
        return
 
    # Initialize a list of empty lists to store nodes at each level
    ans = [[] for i in range(height(root))]
    level = 0
 
    # Intialize the queue for level order traversal
    q = []
    q.append(root)
 
    # Traverse the tree in level order and store nodes at each level in the ans list
    while len(q) > 0:
        n = len(q)
        for i in range(n):
            front_node = q.pop(0)
            ans[level].append(front_node.data)
            if front_node.left:
                q.append(front_node.left)
            if front_node.right:
                q.append(front_node.right)
        level += 1
 
    # Traverse the ans list and print nodes in top-level order and reversed bottom-level order alternatively
    it1 = 0
    it2 = len(ans) - 1
    while it1 < it2:
        # Print nodes in top-level order
        for i in range(len(ans[it1])):
            print(ans[it1][i])
        # Print nodes in reversed bottom-level order
        for i in range(len(ans[it2])-1, -1, -1):
            print(ans[it2][i])
        it1 += 1
        it2 -= 1
    # If there is only one level left, print nodes in top-level order
    if it1 == it2:
        for i in range(len(ans[it1])):
            print(ans[it1][i])
 
 
# Driver program to test the above function
root = Node(25)
root.left = Node(20)
root.right = Node(36)
root.left.left = Node(10)
root.left.right = Node(22)
root.left.left.left = Node(5)
root.left.left.right = Node(12)
root.right.left = Node(30)
root.right.right = Node(40)
root.right.left.left = Node(28)
root.right.right.left = Node(38)
root.right.right.right = Node(48)
 
printBST(root)

C#

// C# Program for the above approach
 
using System;
using System.Collections.Generic;
 
// structure of tree node
class Node {
    public int data;
    public Node left, right;
    public Node(int data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
 
// class to print the nodes of a
// BST in Top Level Order and Reversed
// Bottom Level Order alternatively
class MainClass {
    // function to find the height of binary tree
    public static int Height(Node root)
    {
        if (root == null)
            return 0;
        return Math.Max(Height(root.left),
                        Height(root.right))
            + 1;
    }
 
    public static void PrintBST(Node root)
    {
        // base case
        if (root == null)
            return;
 
        List<List<int> > ans = new List<List<int> >();
        int level = 0;
 
        // initializing the queue for level order traversal
        Queue<Node> q = new Queue<Node>();
        q.Enqueue(root);
 
        while (q.Count != 0) {
            int n = q.Count;
            List<int> currLevel = new List<int>();
            for (int i = 0; i < n; i++) {
                Node front_node = q.Dequeue();
                currLevel.Add(front_node.data);
                if (front_node.left != null)
                    q.Enqueue(front_node.left);
                if (front_node.right != null)
                    q.Enqueue(front_node.right);
            }
            ans.Add(currLevel);
            level++;
        }
 
        int it1 = 0;
        int it2 = ans.Count - 1;
        while (it1 < it2) {
            foreach(int i in ans[it1])
            {
                Console.Write(i + " ");
            }
            for (int i = ans[it2].Count - 1; i >= 0; i--) {
                Console.Write(ans[it2][i] + " ");
            }
            it1++;
            it2--;
        }
        if (it1 == it2) {
            foreach(int i in ans[it1])
            {
                Console.Write(i + " ");
            }
        }
    }
 
    // driver code to test above function
    public static void Main()
    {
        // creating the binary tree
        Node root = new Node(25);
        root.left = new Node(20);
        root.right = new Node(36);
        root.left.left = new Node(10);
        root.left.right = new Node(22);
        root.left.left.left = new Node(5);
        root.left.left.right = new Node(12);
        root.right.left = new Node(30);
        root.right.right = new Node(40);
        root.right.left.left = new Node(28);
        root.right.right.left = new Node(38);
        root.right.right.right = new Node(48);
 
        PrintBST(root);
    }
}
 
// This code is contributed by adityashatmfh

Output

25 48 38 28 12 5 20 36 40 30 22 10

Time Complexity: O(V) where V is the number of vertices in given binary tree.
Auxiliary Space: O(V) due to queue and vector of ans.

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