Clockwise Triangular traversal of a Binary Tree
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> |
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.
Contact Us