Number of Isosceles triangles in a binary tree
Pre-Requisites: Depth First Search | Parent Array Representation
Given a parent array representation of a binary tree, we need to find the number of Isosceles triangles in the binary tree.
Consider a parent array representing a binary tree:
Parent Array:
Given below is the tree representation of the given parent array.
Binary Tree:
There are three types of isosceles triangles which can be found inside a binary tree. These three different types of isosceles triangles can be handled as three different cases.
Case 1: Apex(Vertex against the base sharing equal sides) has two successors(both direct/indirect).
This case can be represented as:
In the given tree, there are 6 such isosceles triangles i.e; (0, 1, 2), (0, 3, 6), (1, 3, 4), (1, 7, 9), (4, 8, 9), (2, 5, 6)
Pseudo Code:
Case 2: Apex has a left successor(direct/indirect) and apex itself is a right successor(direct/indirect) of its parent.
This case can be represented as:
In the given tree, there are 2 such isosceles triangles i.e; (1, 8, 4), (0, 5, 2)
Pseudo Code:
Case 3: Apex has a right successor(direct/indirect) and apex itself is a left successor(direct/indirect) of its parent.
This case can be represented as:
In the given tree, there is 1 such isosceles triangle i.e; (0, 1, 4)
Pseudo Code:
Pseudo Code legend:
left_down[i] -> maximum distance of ith node from its farthest left successor
right_down[i] -> maximum distance of ith node from its farthest right successor
left_up[i] -> maximum distance of ith node from its farthest left predecessor
right_up[i] -> maximum distance of ith node from its farthest right predecessor
Below is the implementation to calculate the number of isosceles triangles present in a given binary tree:
C++
/* C++ program for calculating number of isosceles triangles present in a binary tree */#include <bits/stdc++.h>using namespace std;#define MAX_SZ int(1e5)/* Data Structure used to store Binary Tree in form of Graph */vector<int>* graph;// Data variablesint right_down[MAX_SZ];int left_down[MAX_SZ];int right_up[MAX_SZ];int left_up[MAX_SZ];/* Utility function used to start a DFS traversal over a node */void DFS(int u, int* parent){ if (graph[u].size() != 0) sort(graph[u].begin(), graph[u].end()); if (parent[u] != -1) { if (graph[parent[u]].size() > 1) { /* check if current node is left child of its parent */ if (u == graph[parent[u]][0]) { right_up[u] += right_up[parent[u]] + 1; } // current node is right child of its parent else { left_up[u] += left_up[parent[u]] + 1; } } /* check if current node is left and only child of its parent */ else { right_up[u] += right_up[parent[u]] + 1; } } for (int i = 0; i < graph[u].size(); ++i) { int v = graph[u][i]; // iterating over subtree DFS(v, parent); // left child of current node if (i == 0) { left_down[u] += left_down[v] + 1; } // right child of current node else { right_down[u] += right_down[v] + 1; } }}/* utility function used to generate graph from parent array */int generateGraph(int* parent, int n){ int root; graph = new vector<int>[n]; // Generating graph from parent array for (int i = 0; i < n; ++i) { // check for non-root node if (parent[i] != -1) { /* creating an edge from node with number parent[i] to node with number i */ graph[parent[i]].push_back(i); } // initializing root else { root = i; } // Initializing necessary data variables left_up[i] = 0; right_up[i] = 0; left_down[i] = 0; right_down[i] = 0; } // root of the binary tree return root;}// Driver Functionint main(){ int n = 10; /* Parent array used for storing parent of each node */ int parent[] = { -1, 0, 0, 1, 1, 2, 2, 3, 4, 4 }; /* generateGraph() function generates a graph a returns root of the graph which can be used for starting DFS traversal */ int root = generateGraph(parent, n); // triggering dfs for traversal over graph DFS(root, parent); int count = 0; // Calculation of number of isosceles triangles for (int i = 0; i < n; ++i) { count += min(right_down[i], right_up[i]); count += min(left_down[i], left_up[i]); count += min(left_down[i], right_down[i]); } cout << "Number of isosceles triangles " << "in the given binary tree are " << count; return 0;} |
Java
/* JAVA program for calculating number of isosceles triangles present in a binary tree */import java.io.*;import java.util.*;@SuppressWarnings("unchecked")class Isosceles_triangles { static int MAX_SZ = (int)1e5; /* Data Structure used to store Binary Tree in form of Graph */ static ArrayList<Integer>[] graph; // Data variables static int[] right_down = new int[MAX_SZ]; static int[] left_down = new int[MAX_SZ]; static int[] right_up = new int[MAX_SZ]; static int[] left_up = new int[MAX_SZ]; /* Utility function used to start a DFS traversal over a node */ public static void DFS(int u, int[] parent) { if (graph[u] != null) Collections.sort(graph[u]); if (parent[u] != -1) { if (graph[parent[u]].size() > 1) { /* check if current node is left child of its parent */ if (u == graph[parent[u]].get(0)) { right_up[u] += right_up[parent[u]] + 1; } // current node is right child of its parent else { left_up[u] += left_up[parent[u]] + 1; } } /* check if current node is left and only child of its parent */ else { right_up[u] += right_up[parent[u]] + 1; } } if (graph[u] == null) return; for (int i = 0; i < graph[u].size(); ++i) { int v = graph[u].get(i); // iterating over subtree DFS(v, parent); // left child of current node if (i == 0) { left_down[u] += left_down[v] + 1; } // right child of current node else { right_down[u] += right_down[v] + 1; } } } static int min(Integer a, Integer b) { return (a < b) ? a : b; } /* utility function used to generate graph from parent array */ public static int generateGraph(int[] parent, int n) { int root = -1; graph = (ArrayList<Integer>[]) new ArrayList[n]; // Generating graph from parent array for (int i = 0; i < n; ++i) { // check for non-root node if (parent[i] != -1) { /* creating an edge from node with number parent[i] to node with number i */ if (graph[parent[i]] == null) { graph[parent[i]] = new ArrayList<Integer>(); } graph[parent[i]].add(i); // System.out.println(graph); } // initializing root else { root = i; } // Initializing necessary data variables left_up[i] = 0; right_up[i] = 0; left_down[i] = 0; right_down[i] = 0; } // root of the binary tree return root; } // Driver Function public static void main(String[] args) { int n = 10; /* Parent array used for storing parent of each node */ int[] parent = new int[] { -1, 0, 0, 1, 1, 2, 2, 3, 4, 4 }; /* generateGraph() function generates a graph a returns root of the graph which can be used for starting DFS traversal */ int root = generateGraph(parent, n); // System.exit(0); // triggering dfs for traversal over graph DFS(root, parent); int count = 0; // Calculation of number of isosceles triangles for (int i = 0; i < n; ++i) { count += min(right_down[i], right_up[i]); count += min(left_down[i], left_up[i]); count += min(left_down[i], right_down[i]); } System.out.println("Number of isosceles triangles " + "in the given binary tree are " + Integer.toString(count)); System.exit(0); }} |
Python3
''' Python3 program for calculating number of isosceles triangles present in a binary tree '''MAX_SZ = int(1e5)''' Data Structure used to store Binary Tree in form of Graph '''graph = {}# Data variables right_down = MAX_SZ*[0]left_down = MAX_SZ*[0]right_up = MAX_SZ*[0]left_up = MAX_SZ*[0]''' Utility function used to start a DFS traversal over a node '''def DFS(u, parent): if u in graph: graph[u].sort() if parent[u] != -1: if u in graph and len(graph[parent[u]]) > 1: ''' check if current node is left child of its parent ''' if u == graph[parent[u]][0] : right_up[u] += right_up[parent[u]] + 1 # current node is right child of its parent else: left_up[u] += left_up[parent[u]] + 1 else : ''' check if current node is left and only child of its parent ''' right_up[u] += right_up[parent[u]] + 1 if u in graph: for i in range(0, len(graph[u])): v = graph[u][i] # iterating over subtree DFS(v, parent) # left child of current node if i == 0: left_down[u] += left_down[v] + 1; # right child of current node else: right_down[u] += right_down[v] + 1;''' utility function used to generate graph from parent array '''def generateGraph(parent, n): root = -1 # Generating graph from parent array for i in range(0, n): # check for non-root node if parent[i] != -1: ''' creating an edge from node with number parent[i] to node with number i ''' if parent[i] not in graph: graph[parent[i]] = [i] else : graph[parent[i]].append(i) # initializing root else : root = i # root of the binary tree return root;# Driver Functionif __name__ == '__main__': n = 10 ''' Parent array used for storing parent of each node ''' parent = [-1, 0, 0, 1, 1, 2, 2, 3, 4, 4] ''' generateGraph() function generates a graph a returns root of the graph which can be used for starting DFS traversal ''' root = generateGraph(parent, n) # triggering dfs for traversal over graph DFS(root, parent) count = 0 # Calculation of number of isosceles triangles for i in range(0, n): count += min(right_down[i], right_up[i]) count += min(left_down[i], left_up[i]) count += min(left_down[i], right_down[i]) print("Number of isosceles triangles " + "in the given binary tree are " + str(count)) |
C#
/* C# program for calculating number of isosceles triangles present in a binary tree */using System;using System.Collections.Generic;using System.Linq;class Isosceles_triangles{ static int MAX_SZ = (int)1e5; /* Data Structure used to store Binary Tree in form of Graph */ static List<int>[] graph; // Data variables static int[] right_down = new int[MAX_SZ]; static int[] left_down = new int[MAX_SZ]; static int[] right_up = new int[MAX_SZ]; static int[] left_up = new int[MAX_SZ]; /* Utility function used to start a DFS traversal over a node */ public static void DFS(int u, int[] parent) { if (graph[u] != null) graph[u].Sort(); if (parent[u] != -1) { if (graph[parent[u]].Count > 1) { /* check if current node is left child of its parent */ if (u == graph[parent[u]][0]) { right_up[u] += right_up[parent[u]] + 1; } // current node is right child of its parent else { left_up[u] += left_up[parent[u]] + 1; } } /* check if current node is left and only child of its parent */ else { right_up[u] += right_up[parent[u]] + 1; } } if (graph[u] == null) return; for (int i = 0; i < graph[u].Count; ++i) { int v = graph[u][i]; // iterating over subtree DFS(v, parent); // left child of current node if (i == 0) { left_down[u] += left_down[v] + 1; } // right child of current node else { right_down[u] += right_down[v] + 1; } } } static int min(int a, int b) { return (a < b) ? a : b; } /* utility function used to generate graph from parent array */ public static int generateGraph(int[] parent, int n) { int root = -1; graph = new List<int>[n]; // Generating graph from parent array for (int i = 0; i < n; ++i) { // check for non-root node if (parent[i] != -1) { /* creating an edge from node with number parent[i] to node with number i */ if (graph[parent[i]] == null) { graph[parent[i]] = new List<int>(); } graph[parent[i]].Add(i); // Console.WriteLine(graph); } // initializing root else { root = i; } // Initializing necessary data variables left_up[i] = 0; right_up[i] = 0; left_down[i] = 0; right_down[i] = 0; } // root of the binary tree return root; } // Driver Function public static void Main(String[] args) { int n = 10; /* Parent array used for storing parent of each node */ int[] parent = new int[] { -1, 0, 0, 1, 1, 2, 2, 3, 4, 4 }; /* generateGraph() function generates a graph a returns root of the graph which can be used for starting DFS traversal */ int root = generateGraph(parent, n); // System.exit(0); // triggering dfs for traversal over graph DFS(root, parent); int count = 0; // Calculation of number of isosceles triangles for (int i = 0; i < n; ++i) { count += min(right_down[i], right_up[i]); count += min(left_down[i], left_up[i]); count += min(left_down[i], right_down[i]); } Console.WriteLine("Number of isosceles triangles " + "in the given binary tree are " + count); }}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript program for calculating number of // isosceles triangles present in a binary tree let MAX_SZ = 1e5;// Data Structure used to store // Binary Tree in form of Graph let graph;// Data variableslet right_down = new Array(MAX_SZ);let left_down = new Array(MAX_SZ);let right_up = new Array(MAX_SZ);let left_up = new Array(MAX_SZ);// Utility function used to start// a DFS traversal over a node function DFS(u, parent){ if (graph[u] != null) graph[u].sort(); if (parent[u] != -1) { if (graph[parent[u]].length > 1) { // Check if current node is // left child of its parent if (u == graph[parent[u]][0]) { right_up[u] += right_up[parent[u]] + 1; } // Current node is right child of its parent else { left_up[u] += left_up[parent[u]] + 1; } } // Check if current node is left and // only child of its parent else { right_up[u] += right_up[parent[u]] + 1; } } if (graph[u] == null) return; for(let i = 0; i < graph[u].length; ++i) { let v = graph[u][i]; // Iterating over subtree DFS(v, parent); // left child of current node if (i == 0) { left_down[u] += left_down[v] + 1; } // right child of current node else { right_down[u] += right_down[v] + 1; } }}function min(a, b){ return (a < b) ? a : b;}// Utility function used to generate // graph from parent array function generateGraph(parent, n){ let root = -1; graph = new Array(n); // Generating graph from parent array for(let i = 0; i < n; ++i) { // Check for non-root node if (parent[i] != -1) { // Creating an edge from node with number // parent[i] to node with number i if (graph[parent[i]] == null) { graph[parent[i]] = []; } graph[parent[i]].push(i); // System.out.println(graph); } // Initializing root else { root = i; } // Initializing necessary data variables left_up[i] = 0; right_up[i] = 0; left_down[i] = 0; right_down[i] = 0; } // Root of the binary tree return root;}// Driver codelet n = 10;// Parent array used for storing // parent of each node let parent = [ -1, 0, 0, 1, 1, 2, 2, 3, 4, 4 ];// generateGraph() function generates a graph a // returns root of the graph which can be used for// starting DFS traversal let root = generateGraph(parent, n);// System.exit(0);// Triggering dfs for traversal over graphDFS(root, parent);let count = 0;// Calculation of number of isosceles trianglesfor(let i = 0; i < n; ++i) { count += min(right_down[i], right_up[i]); count += min(left_down[i], left_up[i]); count += min(left_down[i], right_down[i]);}document.write("Number of isosceles triangles " + "in the given binary tree are " + count);// This code is contributed by suresh07</script> |
Output:
Number of isosceles triangles in the given binary tree are 9
Time Complexity : O(n)
Auxiliary Space : O(n)
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