Left-Right traversal of all the levels of Binary tree

Difficulty Level : Medium
Last Updated : 16 Aug, 2021

Given a Binary Tree rooted at node 1, the task is to print the elements in the following defined order.

First, print all elements of the last level in an alternate way such as first you print leftmost element and then rightmost element & continue in this until all elements are traversed of last level.
Now do the same for the rest of the levels.

Examples:

Input:
            1
          /   \
        2      3
      /   \   /
     4     5 6
Output: 4 6 5 2 3 1
Explanation:
First print all elements of the last 
level which will be printed as follows: 4 6 5
Now tree becomes
       1
     /   \
    2     3
    
Now print elements as 2 3
Now the tree becomes: 1

Input:
        1
      /   \
     2     3
Output: 2 3 1

Approach:

Make a bfs call and store all the nodes present at level i int a vector array.
Also keep track of maximum level reached in a bfs call.
Now print the desired pattern starting from max level to 0

Below is the implementation of the above approach:

C++

// C++ implementation
// for the above approach
#include <bits/stdc++.h>
using namespace std;
 
const int sz = 1e5;
int maxLevel = 0;
 
// Adjacency list
// representation of the tree
vector<int> tree[sz + 1];
 
// Boolean array to mark all the
// vertices which are visited
bool vis[sz + 1];
 
// Integer array to store
// the level of each node
int level[sz + 1];
 
// Array of vector where ith index
// stores all the nodes at level i
vector<int> nodes[sz + 1];
 
// Utility function to create an
// edge between two vertices
void addEdge(int a, int b)
{
    // Add a to b's list
    tree[a].push_back(b);
 
    // Add b to a's list
    tree[b].push_back(a);
}
 
// Modified Breadth-First Function
void bfs(int node)
{
  // Create a queue of {child, parent}
  queue<pair<int, int> > qu;
 
  // Push root node in the front of
  // the queue and mark as visited
  qu.push({ node, 0 });
  nodes[0].push_back(node);
  vis[node] = true;
  level[1] = 0;
 
  while (!qu.empty()) {
 
    pair<int, int> p = qu.front();
    // Dequeue a vertex from queue
    qu.pop();
    vis[p.first] = true;
 
    // Get all adjacent vertices of the dequeued
    // vertex s. If any adjacent has not
    // been visited then enqueue it
    for (int child : tree[p.first]) {
        if (!vis[child]) {
            qu.push({ child, p.first });
            level[child] = level[p.first] + 1;
            maxLevel = max(maxLevel, level[child]);
            nodes[level[child]].push_back(child);
        }
    }
  }
}
 
// Function to display
// the pattern
void display()
{
  for (int i = maxLevel; i >= 0; i--) {
    int len = nodes[i].size();
    // Printing all nodes
    // at given level
    for (int j = 0; j < len / 2; j++) {
        cout << nodes[i][j] << " " << nodes[i][len - 1 - j] << " ";
    }
    // If count of nodes
    // at level i is odd
    // print remaining node
    if (len % 2 == 1) {
        cout << nodes[i][len / 2] << " ";
    }
  }
}
 
// Driver code
int main()
{
  // Number of vertices
  int n = 6;
 
  addEdge(1, 2);
  addEdge(1, 3);
  addEdge(2, 4);
  addEdge(2, 5);
  addEdge(3, 6);
 
  // Calling modified bfs function
  bfs(1);
 
  display();
 
  return 0;
}

Java

// Java implementation
// for the above approach
import java.util.*;
 
@SuppressWarnings("unchecked")
 
class GFG{
  
static int sz = 100000;
static int maxLevel = 0;
   
// Adjacency list
// representation of the tree
static ArrayList []tree = new ArrayList[sz + 1];
   
// boolean array to mark all the
// vertices which are visited
static boolean []vis = new boolean[sz + 1];
   
// Integer array to store
// the level of each node
static int []level = new int[sz + 1];
   
// Array of vector where ith index
// stores all the nodes at level i
static ArrayList []nodes = new ArrayList[sz + 1];
   
// Utility function to create an
// edge between two vertices
static void addEdge(int a, int b)
{
     
    // Add a to b's list
    tree[a].add(b);
   
    // Add b to a's list
    tree[b].add(a);
}
 
static class Pair
{
    int Key, Value;
    Pair(int Key, int Value)
    {
        this.Key = Key;
        this.Value = Value;
    }
}
   
// Modified Breadth-Key Function
static void bfs(int node)
{
     
    // Create a queue of {child, parent}
    Queue<Pair> qu = new LinkedList<>();
     
    // Push root node in the front of
    // the queue and mark as visited
    qu.add(new Pair(node, 0));
    nodes[0].add(node);
    vis[node] = true;
    level[1] = 0;
      
    while (qu.size() != 0)
    {
        Pair p = qu.poll();
                                              
        // Dequeue a vertex from queue
        vis[p.Key] = true;
          
        // Get all adjacent vertices of the dequeued
        // vertex s. If any adjacent has not
        // been visited then put it
        for(int child : (ArrayList<Integer>)tree[p.Key])
        {
            if (!vis[child])
            {
                qu.add(new Pair(child, p.Key));
                level[child] = level[p.Key] + 1;
                maxLevel = Math.max(maxLevel,
                                    level[child]);
                nodes[level[child]].add(child);
            }
        }
    }
}
   
// Function to display
// the pattern
static void display()
{
    for(int i = maxLevel; i >= 0; i--)
    {
        int len = nodes[i].size();
         
        // Printing all nodes
        // at given level
        for(int j = 0; j < len / 2; j++)
        {
            System.out.print((int)nodes[i].get(j) + " " +
                             (int)nodes[i].get(len - 1 - j) +
                             " ");
        }
          
        // If count of nodes
        // at level i is odd
        // print remaining node
        if (len % 2 == 1)
        {
            System.out.print(
                (int)nodes[i].get(len / 2) + " ");
        }
    }
}
  
// Driver code
public static void main(String[] args)
{
    for(int i = 0; i < sz + 1; i++)
    {
        tree[i] = new ArrayList();
        nodes[i] = new ArrayList();
        vis[i] = false;
        level[i] = 0;
    }
      
    addEdge(1, 2);
    addEdge(1, 3);
    addEdge(2, 4);
    addEdge(2, 5);
    addEdge(3, 6);
      
    // Calling modified bfs function
    bfs(1);
      
    display();
}
}
 
// This code is contributed by pratham76

Python3

# Python3 implementation
# for the above approach
from collections import deque
 
sz = 10 ** 5
maxLevel = 0
 
# Adjacency list
# representation of the tree
tree = [[] for i in range(sz + 1)]
 
# Boolean array to mark all the
# vertices which are visited
vis = [False] * (sz + 1)
 
# Integer array to store
# the level of each node
level = [0] * (sz + 1)
 
# Array of vector where ith index
# stores all the nodes at level i
nodes = [[] for i in range(sz + 1)]
 
# Utility function to create an
# edge between two vertices
def addEdge(a, b):
     
    global tree
     
    # Add a to b's list
    tree[a].append(b)
 
    # Add b to a's list
    tree[b].append(a)
 
# Modified Breadth-First Function
def bfs(node):
     
    global maxLevel
     
    # Create a queue of {child, parent}
    qu =  deque()
     
    # Push root node in the front of
    # the queue and mark as visited
    qu.append([node, 0 ])
 
    nodes[0].append(node)
    vis[node] = True
    level[1] = 0
 
    while (len(qu) > 0):
 
        p = qu.popleft()
         
        # Dequeue a vertex from queue
        # qu.pop()
        vis[p[0]] = True
 
        # Get all adjacent vertices of the dequeued
        # vertex s. If any adjacent has not
        # been visited then enqueue it
        for child in tree[p[0]]:
            if (not vis[child]):
                qu.append([child, p[0]])
                level[child] = level[p[0]] + 1
                maxLevel = max(maxLevel, level[child])
                nodes[level[child]].append(child)
 
# Function to display
# the pattern
def display():
 
    for i in range(maxLevel, -1, -1):
        lenn = len(nodes[i])
         
        # Printing all nodes
        # at given level
        for j in range(lenn // 2):
            print(nodes[i][j],
                  nodes[i][lenn - 1 - j], end = " ")
             
        # If count of nodes
        # at level i is odd
        # prremaining node
        if (lenn % 2 == 1):
            print(nodes[i][lenn // 2], end = " ")
 
# Driver code
if __name__ == '__main__':
 
    # Number of vertices
    n = 6
 
    addEdge(1, 2)
    addEdge(1, 3)
    addEdge(2, 4)
    addEdge(2, 5)
    addEdge(3, 6)
     
    # Calling modified bfs function
    bfs(1)
 
    display()
 
# This code is contributed by mohit kumar 29

C#

// C# implementation
// for the above approach
using System;
using System.Collections.Generic;
using System.Collections;
 
class GFG{
 
static int sz = 100000;
static int maxLevel = 0;
  
// Adjacency list
// representation of the tree
static ArrayList []tree = new ArrayList[sz + 1];
  
// Boolean array to mark all the
// vertices which are visited
static bool []vis = new bool[sz + 1];
  
// Integer array to store
// the level of each node
static int []level = new int[sz + 1];
  
// Array of vector where ith index
// stores all the nodes at level i
static ArrayList []nodes = new ArrayList[sz + 1];
  
// Utility function to create an
// edge between two vertices
static void addEdge(int a, int b)
{
     
    // Add a to b's list
    tree[a].Add(b);
  
    // Add b to a's list
    tree[b].Add(a);
}
  
// Modified Breadth-First Function
static void bfs(int node)
{
     
    // Create a queue of {child, parent}
    Queue qu = new Queue();
     
    // Push root node in the front of
    // the queue and mark as visited
    qu.Enqueue(new KeyValuePair<int, int>(node, 0));
    nodes[0].Add(node);
    vis[node] = true;
    level[1] = 0;
     
    while(qu.Count != 0)
    {
     
        KeyValuePair<int,
                     int> p = (KeyValuePair<int,
                                            int>)qu.Peek();
                                             
        // Dequeue a vertex from queue
        qu.Dequeue();
        vis[p.Key] = true;
         
        // Get all adjacent vertices of the dequeued
        // vertex s. If any adjacent has not
        // been visited then enqueue it
        foreach(int child in tree[p.Key])
        {
            if (!vis[child])
            {
                qu.Enqueue(new KeyValuePair<int, int>(
                           child, p.Key));
                level[child] = level[p.Key] + 1;
                maxLevel = Math.Max(maxLevel, level[child]);
                nodes[level[child]].Add(child);
            }
        }
    }
}
  
// Function to display
// the pattern
static void display()
{
     
    for(int i = maxLevel; i >= 0; i--)
    {
        int len = nodes[i].Count;
         
        // Printing all nodes
        // at given level
        for(int j = 0; j < len / 2; j++)
        {
            Console.Write((int)nodes[i][j] + " " +
                          (int)nodes[i][len - 1 - j] +
                           " ");
        }
         
        // If count of nodes
        // at level i is odd
        // print remaining node
        if (len % 2 == 1)
        {
            Console.Write((int)nodes[i][len / 2] + " ");
        }
    }
}
 
// Driver code
public static void Main(string[] args)
{
    for(int i = 0; i < sz + 1; i++)
    {
        tree[i] = new ArrayList();
        nodes[i] = new ArrayList();
        vis[i] = false;
        level[i] = 0;
    }
     
    addEdge(1, 2);
    addEdge(1, 3);
    addEdge(2, 4);
    addEdge(2, 5);
    addEdge(3, 6);
     
    // Calling modified bfs function
    bfs(1);
     
    display();
}
}
 
// This code is contributed by rutvik_56

Javascript

<script>
 
// JavaScript implementation
// for the above approach
 
let sz = 100000;
let maxLevel = 0;
 
// Adjacency list
// representation of the tree
let tree = new Array(sz + 1);
 
// boolean array to mark all the
// vertices which are visited
let vis = new Array(sz + 1);
 
// Integer array to store
// the level of each node
let level = new Array(sz + 1);
 
// Array of vector where ith index
// stores all the nodes at level i
let nodes = new Array(sz + 1);
 
// Utility function to create an
// edge between two vertices
function addEdge(a,b)
{
    // Add a to b's list
    tree[a].push(b);
    
    // Add b to a's list
    tree[b].push(a);
}
 
// Modified Breadth-Key Function
function bfs(node)
{
    let qu=[];
    // Push root node in the front of
    // the queue and mark as visited
    qu.push([node, 0]);
    nodes[0].push(node);
    vis[node] = true;
    level[1] = 0;
       
    while (qu.length != 0)
    {
        let p = qu.shift();
                                               
        // Dequeue a vertex from queue
        vis[p[0]] = true;
           
        // Get all adjacent vertices of the dequeued
        // vertex s. If any adjacent has not
        // been visited then put it
        for(let child=0;child<tree[p[0]].length;child++)
        {
            if (!vis[tree[p[0]][child]])
            {
                 
                qu.push([tree[p[0]][child], p[0]]);
                level[tree[p[0]][child]] = level[p[0]] + 1;
                maxLevel = Math.max(maxLevel,
                                    level[tree[p[0]][child]]);
                                     
                 
                nodes[level[tree[p[0]][child]]].
                push(tree[p[0]][child]);
                 
            }
        }
    }
}
 
// Function to display
// the pattern
function display()
{
     
    for(let i = maxLevel; i >= 0; i--)
    {
        let len = nodes[i].length;  
        // Printing all nodes
        // at given level
        for(let j = 0; j < Math.floor(len / 2); j++)
        {
            document.write(nodes[i][j] + " " +
                             nodes[i][len - 1 - j] +
                             " ");
        }
           
        // If count of nodes
        // at level i is odd
        // print remaining node
        if (len % 2 == 1)
        {
            document.write(
                nodes[i][Math.floor(len / 2)] + " ");
        }
    }
}
 
// Driver code
for(let i = 0; i < sz + 1; i++)
    {
        tree[i] = [];
        nodes[i] = [];
        vis[i] = false;
        level[i] = 0;
    }
       
    addEdge(1, 2);
    addEdge(1, 3);
    addEdge(2, 4);
    addEdge(2, 5);
    addEdge(3, 6);
       
    // Calling modified bfs function
    bfs(1);
       
    display();
 
 
// This code is contributed by patel2127
 
</script>

Output:

4 6 5 2 3 1

Time Complexity: O(V + E), where V is total number of vertices and E is total number of edges.
Auxiliary Space: O(V).

My Personal Notes

Related Articles

Table of Contents

Left-Right traversal of all the levels of Binary tree

C++

Java

Python3

C#

Javascript

Start Your Coding Journey Now!