Open In App

Clockwise Triangular traversal of a Binary Tree

Improve
Improve
Like Article
Like
Save
Share
Report

Given a Complete Binary Tree, the task is to print the elements in the Clockwise traversal order.
Clockwise Traversal of a tree is defined as: 
 

For the above binary tree, the Clockwise Triangular traversal will be 
0, 2, 6, 14, 13, 12, 11, 10, 9, 8, 7, 3, 1, 5, 4 
 

Examples:

Input:
          1
      /       \
     2         3
   /   \      /   \
  4     5    6    7
 / \    /\
8   9  10 11     
Output: 1, 3, 7, 11, 10, 9, 8, 4, 2, 6, 5

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

Approach: 
Create a vector tree[] where tree[i] will store all the nodes of the tree at level i. Take an integer k which keeps track which level we are traversing other integer and cycle in which keep tracks how many cycles have been completed. Now, start printing the nodes the rightmost remaining node which has not been traversed yet & keep moving down until you reach down to the last level which has not been traversed now print this level from right to left, then move print leftmost remaining leftmost element of each level starting from last level to moving to the uppermost level whose elements has all not been traversed yet, now again do the same thing until all elements have not been traversed.

Below is the implementation of the above approach:

C++




// C++ program for the
// above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to create an
// edge between two vertices
void addEdge(int a, int b, vector<int> tree[])
{
 
    // Add a to b's list
    tree[a].push_back(b);
 
    // Add b to a's list
    tree[b].push_back(a);
}
 
// Function to create
// complete binary tree
void createTree(int n, vector<int> tree[])
{
    for (int i = 1;; i++) {
        // Adding edge to
        // a binary tree
        int c = 0;
        if (2 * i <= n) {
            addEdge(i, 2 * i, tree);
            c++;
        }
        if (2 * i + 1 <= n) {
            addEdge(i, 2 * i + 1, tree);
            c++;
        }
 
        if (c == 0)
            break;
    }
}
 
// Modified Breadth-First-Search Function
void bfs(int node, vector<int> tree[], bool vis[],
         int level[], vector<int> nodes[], int& maxLevel)
{
 
    // 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(vector<int> nodes[], int maxLevel)
{
 
    // k represents the level no.
    // cycle represents how many
    // cycles has been completed
    int k = 0, cycle = 0;
 
    // While there are nodes
    // left to traverse
    while (cycle - 1 <= maxLevel / 2) {
 
        // Traversing rightmost element
        // in each cycle as we move down
        while (k < maxLevel - cycle) {
            int j = nodes[k].size() - 1;
            cout << nodes[k][j - cycle] << " ";
            k++;
        }
 
        // Traversing each element of remaining
        // last level from right to left
        if (k == maxLevel - cycle) {
            int j = nodes[k].size() - 1;
            for (j -= cycle; j >= cycle; j--)
                cout << nodes[k][j] << " ";
        }
        k--;
 
        // Traversing leftmost remaining element
        // in each cycle as we move up
        while (k > cycle) {
            cout << nodes[k][cycle] << " ";
            k--;
        }
 
        // No of cycles
        // completed
        cycle++;
 
        // updating from which level to
        // start new cycle
        k = cycle + 1;
    }
}
 
// Driver code
int main()
{
 
    // Number of vertices
    int n = 12;
 
    const int sz = 1e5;
    int maxLevel = 0;
 
    vector<int> tree[sz + 1];
    bool vis[sz + 1];
    int level[sz + 1];
 
    vector<int> nodes[sz + 1];
 
    createTree(n, tree);
 
    bfs(1, tree, vis, level, nodes, maxLevel);
 
    display(nodes, maxLevel);
 
    return 0;
}


Java




// Java code for the above approach
import java.util.*;
 
class Main {
    // Function to create an
    // edge between two vertices
    static void addEdge(int a, int b, List<Integer>[] tree)
    {
        tree[a].add(b);
        tree[b].add(a);
    }
 
    // Function to create
    // complete binary tree
 
    static void createTree(int n, List<Integer>[] tree)
    {
        for (int i = 1;; i++) {
            int c = 0;
            // Adding edge to
            // a binary tree
            if (2 * i <= n) {
                addEdge(i, 2 * i, tree);
                c++;
            }
            if (2 * i + 1 <= n) {
                addEdge(i, 2 * i + 1, tree);
                c++;
            }
 
            if (c == 0)
                break;
        }
    }
    // Modified Breadth-First Function
    static void bfs(int node, List<Integer>[] tree,
                    boolean[] vis, int[] level,
                    List<Integer>[] nodes, int[] maxLevel)
    {
        // Create a queue of
        // {child, parent}
        Queue<int[]> qu = new LinkedList<>();
        // Push root node in the front of
        // the queue and mark as visited
        qu.add(new int[] { node, 0 });
        nodes[0].add(node);
        vis[node] = true;
        level[1] = 0;
 
        while (!qu.isEmpty()) {
            int[] p = qu.poll();
            vis[p[0]] = true;
            for (int child : tree[p[0]]) {
                if (!vis[child]) {
                    qu.add(new int[] { child, p[0] });
                    level[child] = level[p[0]] + 1;
                    maxLevel[0] = Math.max(maxLevel[0],
                                           level[child]);
                    nodes[level[child]].add(child);
                }
            }
        }
    }
 
    static void display(List<Integer>[] nodes, int maxLevel)
    {
        int k = 0, cycle = 0;
        while (cycle - 1 <= maxLevel / 2) {
            while (k < maxLevel - cycle) {
                int j = nodes[k].size() - 1;
                System.out.print(nodes[k].get(j - cycle)
                                 + " ");
                k++;
            }
            if (k == maxLevel - cycle) {
                int j = nodes[k].size() - 1;
                for (j -= cycle; j >= cycle; j--)
                    System.out.print(nodes[k].get(j) + " ");
            }
            k--;
            while (k > cycle) {
                System.out.print(nodes[k].get(cycle) + " ");
                k--;
            }
            cycle++;
            k = cycle + 1;
        }
    }
 
    public static void main(String[] args)
    {
        int n = 12;
        List<Integer>[] tree = new List[100000];
        for (int i = 0; i < tree.length; i++) {
            tree[i] = new ArrayList<>();
        }
        boolean[] vis = new boolean[100000];
        int[] level = new int[100000];
        List<Integer>[] nodes = new List[100000];
        for (int i = 0; i < nodes.length; i++) {
            nodes[i] = new ArrayList<>();
        }
        int[] maxLevel = new int[1];
 
        createTree(n, tree);
        bfs(1, tree, vis, level, nodes, maxLevel);
        display(nodes, maxLevel[0]);
    }
}
// This code is contributed by Potta Lokesh


Python3




# Python3 program for the
# above approach
  
# 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); 
 
# Function to create
# complete binary tree
def createTree(n):
     
    i = 1   
    while True:
     
        # Adding edge to
        # a binary tree
        c = 0;
        if (2 * i <= n):
            addEdge(i, 2 * i);
            c += 1;
         
        if (2 * i + 1 <= n):
            addEdge(i, 2 * i + 1);
            c += 1       
  
        if (c == 0):
            break;
         
        i += 1   
 
# Modified Breadth-First
# Function
def bfs(node, maxLevel):
  
    # Create a queue of
    # {child, parent}
    qu = []
  
    # 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[0];
  
        # Dequeue a vertex
        # from queue
        qu.pop(0);
        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);
                 
    return maxLevel
 
# Function to display
# the pattern
def display(maxLevel):
  
    # k represents the level no.
    # cycle represents how many
    # cycles has been completed
    k = 0
    cycle = 0;
  
    # While there are nodes
    # left to traverse
    while (cycle - 1 <= maxLevel // 2):
  
        # Traversing rightmost element
        # in each cycle as we move down
        while(k < maxLevel - cycle):
            j = len(nodes[k]) - 1;
            print(nodes[k][j - cycle],
                  end = ' ')
            k += 1
  
        # Traversing each element of
        # remaining last level from right
        # to left
        if (k == maxLevel - cycle):
            j = len(nodes[k]) - 1 - cycle;
            while(j >= cycle):               
                print(nodes[k][j],
                      end = ' ')
                j -= 1               
        k -= 1
  
        # Traversing leftmost remaining
        # element in each cycle as we
        # move up
        while (k > cycle):
            print(nodes[k][cycle],
                  end = ' ')
            k -= 1
  
        # No of cycles
        # completed
        cycle += 1
  
        # updating from which
        # level to start new cycle
        k = cycle + 1;   
 
# Driver code
if __name__=="__main__":
     
    # Number of vertices
    n = 12;
  
    sz = 100005;
    maxLevel = 0;   
    tree = [[] for i in range(sz + 1)]   
    vis = [False for i in range(sz + 1)]
    level = [0 for i in range(sz + 1)]
    nodes = [[] for i in range(sz + 1)]
    createTree(n);
    maxLevel = bfs(1, maxLevel);
    display(maxLevel);
 
# This code is contributed by Rutvik_56


C#




// C# program for the
// above approach
using System;
using System.Collections.Generic;
 
class MainClass {
   
  // Function to create an edge between two vertices
  static void AddEdge(List<int>[] tree, int a, int b)
  {
    // Add a to b's list
    tree[a].Add(b);
 
    // Add b to a's list
    tree[b].Add(a);
  }
 
  // Function to create complete binary tree
  static void CreateTree(List<int>[] tree, int n)
  {
    int i = 1;
 
    while (true) {
       
      // Adding edge to a binary tree
      int c = 0;
 
      if (2 * i <= n) {
        AddEdge(tree, i, 2 * i);
        c++;
      }
 
      if (2 * i + 1 <= n) {
        AddEdge(tree, i, 2 * i + 1);
        c++;
      }
 
      if (c == 0) {
        break;
      }
 
      i++;
    }
  }
 
  // Modified Breadth-First Function
  static int BFS(List<int>[] tree, List<int>[] nodes,
                 bool[] vis, int[] level, int node,
                 int maxLevel)
  {
     
    // Create a queue of {child, parent}
    Queue<int[]> qu = new Queue<int[]>();
 
    // Push root node in the front of the queue and mark
    // as visited
    qu.Enqueue(new int[] { node, 0 });
    nodes[0].Add(node);
    vis[node] = true;
    level[1] = 0;
 
    while (qu.Count > 0) {
      int[] p = qu.Dequeue();
 
      // 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[0]])
      {
        if (!vis[child]) {
          qu.Enqueue(new int[] { child, p[0] });
          level[child] = level[p[0]] + 1;
          maxLevel
            = Math.Max(maxLevel, level[child]);
          nodes[level[child]].Add(child);
          vis[child] = true;
        }
      }
    }
 
    return maxLevel;
  }
 
  // Function to display the pattern
  static void Display(List<int>[] nodes, int maxLevel)
  {
     
    // k represents the level no.
    // cycle represents how many cycles has been
    // completed
    int k = 0;
    int cycle = 0;
 
    // While there are nodes left to traverse
    while (cycle - 1 <= maxLevel / 2) {
      // Traversing rightmost element in each cycle as
      // we move down
      while (k < maxLevel - cycle) {
        int j = nodes[k].Count - 1;
        Console.Write("{0} ", nodes[k][j - cycle]);
        k++;
      }
 
      // Traversing each element of remaining last
      // level from right to left
      if (k == maxLevel - cycle) {
        int j = nodes[k].Count - 1 - cycle;
 
        while (j >= cycle) {
          Console.Write("{0} ", nodes[k][j]);
          j--;
        }
      }
 
      k--;
 
      // Traversing leftmost remaining element in each
      // cycle as we move up
      while (k > cycle) {
        Console.Write("{0} ", nodes[k][cycle]);
        k--;
      }
 
      // No of cycles completed
 
      cycle++;
 
      // Updating from which
 
      // level to start new cycle
      k = cycle + 1;
    }
  }
 
  // Driver code
  static void Main(string[] args)
  {
    // Number of vertices
    int n = 12;
 
    int sz = 100005;
    int maxLevel = 0;
    List<int>[] tree = new List<int>[ sz + 1 ];
    bool[] vis = new bool[sz + 1];
    int[] level = new int[sz + 1];
    List<int>[] nodes = new List<int>[ sz + 1 ];
    for (int i = 0; i < sz + 1; i++) {
      tree[i] = new List<int>();
      vis[i] = false;
      level[i] = 0;
      nodes[i] = new List<int>();
    }
 
    CreateTree(tree, n);
    maxLevel
      = BFS(tree, nodes, vis, level, 1, maxLevel);
    Display(nodes, maxLevel);
  }
}
 
// This code is contributed by rutikbhosale


Javascript




<script>
 
    // JavaScript program for the above approach
     
    let sz = 1e5;
    let tree = new Array(sz + 1);
    let nodes = new Array(sz + 1);
    let vis = new Array(sz + 1);
    let level = new Array(sz + 1);
     
    // 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);
    }
 
    // Function to create
    // complete binary tree
    function createTree(n)
    {
        let i = 1;
        while(true) {
            // Adding edge to
            // a binary tree
            let c = 0;
            if (2 * i <= n) {
                addEdge(i, 2 * i);
                c++;
            }
            if (2 * i + 1 <= n) {
                addEdge(i, 2 * i + 1);
                c++;
            }
 
            if (c == 0)
                break;
              i+=1;
        }
    }
 
    // Modified Breadth-First Function
    function bfs(node, maxLevel)
    {
 
        // 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]);
                }
            }
        }
        return maxLevel;
    }
 
    // Function to display the pattern
    function display(maxLevel)
    {
 
        // k represents the level no.
        // cycle represents how many
        // cycles has been completed
        let k = 0, cycle = 0;
 
        // While there are nodes
        // left to traverse
        while (cycle - 1 <= parseInt(maxLevel / 2, 10)) {
 
            // Traversing rightmost element
            // in each cycle as we move down
            while (k < maxLevel - cycle) {
                let j = nodes[k].length - 1;
                document.write(nodes[k][j - cycle] + " ");
                k++;
            }
 
            // Traversing each element of remaining
            // last level from right to left
            if (k == maxLevel - cycle) {
                let j = nodes[k].length - 1;
                for (j -= cycle; j >= cycle; j--)
                    document.write(nodes[k][j] + " ");
            }
            k--;
 
            // Traversing leftmost remaining element
            // in each cycle as we move up
            while (k > cycle) {
                document.write(nodes[k][cycle] + " ");
                k--;
            }
 
            // No of cycles
            // completed
            cycle++;
 
            // updating from which level to
            // start new cycle
            k = cycle + 1;
        }
    }
     
    // Number of vertices
    let n = 12;
     
    for(let i = 0; i < sz + 1; i++)
    {
        tree[i] = [];
        vis[i] = false;
        level[i] = 0;
        nodes[i] = [];
    }
     
    let maxLevel = 0;
  
    createTree(n);
  
    maxLevel = bfs(1, maxLevel);
  
    display(maxLevel);
 
</script>


Output: 

1 3 7 12 11 10 9 8 4 2 6 5

 

Time Complexity: O(n)

Space Complexity: O(n) as the program uses several arrays and vectors to store information about the tree and its traversal.



Last Updated : 29 Mar, 2023
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads