Sideways traversal of a Complete Binary Tree

Last Updated : 07 Feb, 2022

Given a Complete Binary Tree, the task is to print the elements in the following pattern. Let’s consider the tree to be:

The tree is traversed in the following way:

The output for the above tree is:

1 3 7 11 10 9 8 4 5 6 2

Approach: The idea is to use the modified breadth first search function to store all the nodes at every level in an array of vector. Along with it, the maximum level up to which the tree needs to be traversed is also stored in a variable. After this precomputation task, the following steps are followed to get the required answer:

Create a vector tree[] where tree[i] will store all the nodes of the tree at the level i.
Take an integer variable k which keeps the track of the level number that is being traversed and another integer variable path which keeps the track of the number of cycles that have been completed. A flag variable is also created to keep the track of the direction in which the tree is being traversed.
Now, start printing the rightmost nodes at each level until the maximum level is reached.
Since the maximum level is reached, the direction has to be changed. In the last level, print elements from rightmost to left. And the value of maxLevel variable has to be decremented.
As the tree is being traversed from the lower level to the upper level, the rightmost elements are printed. Since in the next iteration, the maxlevel value has been changed, it makes sure that already visited nodes in the last level are not traversed again.

Below is the implementation of the above approach:

C++

// C++ program to print sideways
// traversal of complete binary tree
 
#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);
            }
        }
    }
}
 
// Utility Function to display the pattern
void display()
{
    // k represents the level no.
    // cycle represents how many
    // cycles has been completed
    int k = 0, path = 0;
    int condn = (maxLevel) / 2 + 1;
    bool flag = true;
 
    // While there are nodes left to traverse
    while (condn--) {
 
        if (flag) {
 
            // Traversing whole level from
            // left to right
            int j = nodes[k].size() - 1;
            for (j = 0; j < nodes[k].size() - path; j++)
                cout << nodes[k][j] << " ";
 
            // Moving to new level
            k++;
 
            // Traversing rightmost unvisited
            // element  in path as we
            // move up to down
            while (k < maxLevel) {
 
                j = nodes[k].size() - 1;
                cout << nodes[k][j - path] << " ";
                k++;
            }
 
            j = nodes[k].size() - 1;
            if (k > path)
                for (j -= path; j >= 0; j--)
                    cout << nodes[k][j] << " ";
 
            // Setting value of new maximum
            // level upto which we have to traverse
            // next time
            maxLevel--;
 
            // Updating from which level to
            // start new path
            k--;
            path++;
 
            flag = !flag;
        }
        else {
 
            // Traversing each element of remaining
            // last level from left to right
            int j = nodes[k].size() - 1;
            for (j = 0; j < nodes[k].size() - path; j++)
                cout << nodes[k][j] << " ";
 
            // Decrementing value of Max level
            maxLevel--;
 
            k--;
 
            // Traversing rightmost unvisited
            // element  in path as we
            // move down to up
            while (k > path) {
 
                int j = nodes[k].size() - 1;
                cout << nodes[k][j - path] << " ";
                k--;
            }
 
            j = nodes[k].size() - 1;
 
            if (k == path)
                for (j -= path; j >= 0; j--)
                    cout << nodes[k][j] << " ";
 
            path++;
 
            // Updating the level number from which
            // a new cycle has to be started
            k++;
            flag = !flag;
        }
    }
}
 
// Driver code
int main()
{
 
    // Initialising  the above mentioned
    // complete binary tree
    for (int i = 1; i <= 5; i++) {
 
        // Adding edge to a binary tree
        addEdge(i, 2 * i);
        addEdge(i, 2 * i + 1);
    }
 
    // Calling modified bfs function
    bfs(1);
 
    display();
 
    return 0;
}

Java

// Java program to print sideways
// traversal of complete binary tree
import java.util.*;
 
class GFG
{
     
static class pair
{ 
    int first, second; 
    public pair(int first, int second) 
    { 
        this.first = first; 
        this.second = second; 
    } 
} 
static int sz = (int) 1e5;
static int maxLevel = 0;
 
// Adjacency list representation
// of the tree
static Vector<Integer> []tree = new Vector[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 Vector<Integer> []nodes = new Vector[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<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.isEmpty()) {
 
        pair p = qu.peek();
 
        // Dequeue a vertex from queue
        qu.remove();
        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.add(new pair( child, p.first ));
                level[child] = level[p.first] + 1;
                maxLevel = Math.max(maxLevel, level[child]);
                nodes[level[child]].add(child);
            }
        }
    }
}
 
// Utility Function to display the pattern
static void display()
{
    // k represents the level no.
    // cycle represents how many
    // cycles has been completed
    int k = 0, path = 0;
    int condn = (maxLevel) / 2 + 1;
    boolean flag = true;
 
    // While there are nodes left to traverse
    while (condn-- > 0) {
 
        if (flag) {
 
            // Traversing whole level from
            // left to right
            int j = nodes[k].size() - 1;
            for (j = 0; j < nodes[k].size() - path; j++)
                System.out.print(nodes[k].get(j)+ " ");
 
            // Moving to new level
            k++;
 
            // Traversing rightmost unvisited
            // element in path as we
            // move up to down
            while (k < maxLevel) {
 
                j = nodes[k].size() - 1;
                System.out.print(nodes[k].get(j - path)+ " ");
                k++;
            }
 
            j = nodes[k].size() - 1;
            if (k > path)
                for (j -= path; j >= 0; j--)
                    System.out.print(nodes[k].get(j)+ " ");
 
            // Setting value of new maximum
            // level upto which we have to traverse
            // next time
            maxLevel--;
 
            // Updating from which level to
            // start new path
            k--;
            path++;
 
            flag = !flag;
        }
        else {
 
            // Traversing each element of remaining
            // last level from left to right
            int j = nodes[k].size() - 1;
            for (j = 0; j < nodes[k].size() - path; j++)
                System.out.print(nodes[k].get(j)+ " ");
 
            // Decrementing value of Max level
            maxLevel--;
 
            k--;
 
            // Traversing rightmost unvisited
            // element in path as we
            // move down to up
            while (k > path) {
 
                int c = nodes[k].size() - 1;
                System.out.print(nodes[k].get(c - path)+ " ");
                k--;
            }
 
            j = nodes[k].size() - 1;
 
            if (k == path)
                for (j -= path; j >= 0; j--)
                    System.out.print(nodes[k].get(j)+ " ");
 
            path++;
 
            // Updating the level number from which
            // a new cycle has to be started
            k++;
            flag = !flag;
        }
    }
}
 
// Driver code
public static void main(String[] args)
{
 
    for (int i = 0; i < tree.length; i++) {
        tree[i] = new Vector<>();
        nodes[i] = new Vector<>();
    }
     
    // Initialising the above mentioned
    // complete binary tree
    for (int i = 1; i <= 5; i++) {
 
        // Adding edge to a binary tree
        addEdge(i, 2 * i);
        addEdge(i, 2 * i + 1);
    }
 
    // Calling modified bfs function
    bfs(1);
 
    display();
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to print sideways
# traversal of complete binary tree
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):
 
    # 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
        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 (vis[child] == False):
                qu.append([child, p[0]])
                level[child] = level[p[0]] + 1
                maxLevel = max(maxLevel, level[child])
                nodes[level[child]].append(child)
 
# Utility Function to display the pattern
def display():
    global maxLevel
     
    # k represents the level no.
    # cycle represents how many
    # cycles has been completed
    k = 0
    path = 0
    condn = (maxLevel) // 2 + 1
    flag = True
 
    # While there are nodes left to traverse
    while (condn):
 
        if (flag):
 
            # Traversing whole level from
            # left to right
            j = len(nodes[k]) - 1
            for j in range(len(nodes[k])- path):
                print(nodes[k][j],end=" ")
 
            # Moving to new level
            k += 1
 
            # Traversing rightmost unvisited
            # element in path as we
            # move up to down
            while (k < maxLevel):
 
                j = len(nodes[k]) - 1
                print(nodes[k][j - path], end=" ")
                k += 1
 
            j = len(nodes[k]) - 1
            if (k > path):
                while j >= 0:
                    j -= path
                    print(nodes[k][j], end=" ")
                    j -= 1
 
            # Setting value of new maximum
            # level upto which we have to traverse
            # next time
            maxLevel -= 1
 
            # Updating from which level to
            # start new path
            k -= 1
            path += 1
 
            flag = not flag
        else:
 
            # Traversing each element of remaining
            # last level from left to right
            j = len(nodes[k]) - 1
            for j in range(len(nodes[k]) - path):
                print(nodes[k][j], end=" ")
 
            # Decrementing value of Max level
            maxLevel -= 1
 
            k -= 1
 
            # Traversing rightmost unvisited
            # element in path as we
            # move down to up
            while (k > path):
 
                j = len(nodes[k]) - 1
                print(nodes[k][j - path], end=" ")
                k -= 1
 
            j = len(nodes[k]) - 1
 
            if (k == path):
                while j >= 0:
                    j -= path
                    print(nodes[k][j],end=" ")
                    j -= 1
 
            path += 1
 
            # Updating the level number from which
            # a new cycle has to be started
            k += 1
            flag = not flag
        condn -= 1
 
# Driver code
if __name__ == '__main__':
 
    # Initialising the above mentioned
    # complete binary tree
    for i in range(1,6):
 
        # Adding edge to a binary tree
        addEdge(i, 2 * i)
        addEdge(i, 2 * i + 1)
 
    # Calling modified bfs function
    bfs(1)
 
    display()
 
# This code is contributed by mohit kumar 29

C#

// C# program to print sideways
// traversal of complete binary tree
using System;
using System.Collections.Generic;
 
class GFG
{
      
class pair
{ 
    public int first, second; 
    public pair(int first, int second) 
    { 
        this.first = first; 
        this.second = second; 
    } 
} 
static int sz = (int) 1e5;
static int maxLevel = 0;
  
// Adjacency list representation
// of the tree
static List<int> []tree = new List<int>[sz + 1];
  
// Boolean array to mark all the
// vertices which are visited
static bool []vis = new bool[sz + 1];
  
// int 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 List<int> []nodes = new List<int>[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<pair> qu = new Queue<pair>();
  
    // Push root node in the front of
    // the queue and mark as visited
    qu.Enqueue(new pair( node, 0 ));
    nodes[0].Add(node);
    vis[node] = true;
    level[1] = 0;
  
    while (qu.Count != 0) {
  
        pair p = qu.Peek();
  
        // Dequeue a vertex from queue
        qu.Dequeue();
        vis[p.first] = 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.first]) {
  
            if (!vis[child]) {
                qu.Enqueue(new pair( child, p.first ));
                level[child] = level[p.first] + 1;
                maxLevel = Math.Max(maxLevel, level[child]);
                nodes[level[child]].Add(child);
            }
        }
    }
}
  
// Utility Function to display the pattern
static void display()
{
    // k represents the level no.
    // cycle represents how many
    // cycles has been completed
    int k = 0, path = 0;
    int condn = (maxLevel) / 2 + 1;
    bool flag = true;
  
    // While there are nodes left to traverse
    while (condn-- > 0) {
  
        if (flag) {
  
            // Traversing whole level from
            // left to right
            int j = nodes[k].Count - 1;
            for (j = 0; j < nodes[k].Count - path; j++)
                Console.Write(nodes[k][j]+ " ");
  
            // Moving to new level
            k++;
  
            // Traversing rightmost unvisited
            // element in path as we
            // move up to down
            while (k < maxLevel) {
  
                j = nodes[k].Count - 1;
                Console.Write(nodes[k][j - path]+ " ");
                k++;
            }
  
            j = nodes[k].Count - 1;
            if (k > path)
                for (j -= path; j >= 0; j--)
                    Console.Write(nodes[k][j]+ " ");
  
            // Setting value of new maximum
            // level upto which we have to traverse
            // next time
            maxLevel--;
  
            // Updating from which level to
            // start new path
            k--;
            path++;
  
            flag = !flag;
        }
        else {
  
            // Traversing each element of remaining
            // last level from left to right
            int j = nodes[k].Count - 1;
            for (j = 0; j < nodes[k].Count - path; j++)
                Console.Write(nodes[k][j]+ " ");
  
            // Decrementing value of Max level
            maxLevel--;
  
            k--;
  
            // Traversing rightmost unvisited
            // element in path as we
            // move down to up
            while (k > path) {
  
                int c = nodes[k].Count - 1;
                Console.Write(nodes[k]+ " ");
                k--;
            }
  
            j = nodes[k].Count - 1;
  
            if (k == path)
                for (j -= path; j >= 0; j--)
                    Console.Write(nodes[k][j]+ " ");
  
            path++;
  
            // Updating the level number from which
            // a new cycle has to be started
            k++;
            flag = !flag;
        }
    }
}
  
// Driver code
public static void Main(String[] args)
{
  
    for (int i = 0; i < tree.Length; i++) {
        tree[i] = new List<int>();
        nodes[i] = new List<int>();
    }
      
    // Initialising the above mentioned
    // complete binary tree
    for (int i = 1; i <= 5; i++) {
  
        // Adding edge to a binary tree
        addEdge(i, 2 * i);
        addEdge(i, 2 * i + 1);
    }
  
    // Calling modified bfs function
    bfs(1);
  
    display();
}
}
 
// This code contributed by PrinciRaj1992

Javascript

<script>
// Javascript program to print sideways
// traversal of complete binary tree
 
let sz =  1e5;
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-First Function
function bfs(node)
{
    // Create a queue of {child, parent}
    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[0];
  
        // Dequeue a vertex from queue
        qu.shift();
        vis[p[0]] = true;
  
        // Get all adjacent vertices of the dequeued
        // vertex s. If any adjacent has not
        // been visited then enqueue 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]);
            }
        }
    }
}
 
// Utility Function to display the pattern
function display()
{
    // k represents the level no.
    // cycle represents how many
    // cycles has been completed
    let k = 0, path = 0;
    let condn = Math.floor((maxLevel) / 2) + 1;
    let flag = true;
  
    // While there are nodes left to traverse
    while (condn-- > 0) {
  
        if (flag) {
  
            // Traversing whole level from
            // left to right
            let j = nodes[k].length - 1;
            for (j = 0; j < nodes[k].length - path; j++)
                document.write(nodes[k][j]+ " ");
  
            // Moving to new level
            k++;
  
            // Traversing rightmost unvisited
            // element in path as we
            // move up to down
            while (k < maxLevel) {
  
                j = nodes[k].length - 1;
                document.write(nodes[k][j - path]+ " ");
                k++;
            }
  
            j = nodes[k].length - 1;
            if (k > path)
                for (j -= path; j >= 0; j--)
                    document.write(nodes[k][j]+ " ");
  
            // Setting value of new maximum
            // level upto which we have to traverse
            // next time
            maxLevel--;
  
            // Updating from which level to
            // start new path
            k--;
            path++;
  
            flag = !flag;
        }
        else {
  
            // Traversing each element of remaining
            // last level from left to right
            let j = nodes[k].length - 1;
            for (j = 0; j < nodes[k].length - path; j++)
                document.write(nodes[k][j]+ " ");
  
            // Decrementing value of Max level
            maxLevel--;
  
            k--;
  
            // Traversing rightmost unvisited
            // element in path as we
            // move down to up
            while (k > path) {
  
                let c = nodes[k].length - 1;
                document.write(nodes[k]+ " ");
                k--;
            }
  
            j = nodes[k].length - 1;
  
            if (k == path)
                for (j -= path; j >= 0; j--)
                    document.write(nodes[k][j]+ " ");
  
            path++;
  
            // Updating the level number from which
            // a new cycle has to be started
            k++;
            flag = !flag;
        }
    }
}
 
// Driver code
for (let i = 0; i < tree.length; i++) {
        tree[i] = [];
        nodes[i] = [];
        vis[i]=false;
        level[i]=0;
    }
      
    // Initialising the above mentioned
    // complete binary tree
    for (let i = 1; i <= 5; i++) {
  
        // Adding edge to a binary tree
        addEdge(i, 2 * i);
        addEdge(i, 2 * i + 1);
    }
  
    // Calling modified bfs function
    bfs(1);
  
    display();
 
 
// This code is contributed by unknown2108
</script>

Output:

1 3 7 11 10 9 8 4 5 6 2

Suggest improvement

Least Common Ancestor of any number of nodes in Binary Tree

CamelCase Pattern Matching

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Sideways traversal of a Complete Binary Tree

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