Given three vertices U, V and R of a binary tree, the task is to check whether R lies in the path between U and V. If it is not present in the path then print No otherwise print Yes.
Examples:
Input: U = 4, V = 6, R = 2
Output: Yes
Path 4 -> 2 -> 1 -> 3 -> 6 contains 2
Input: U = 4, V = 6, R = 5
Output: No
Path 4 -> 2 -> 1 -> 3 -> 6 does not contain 5
Approach: The idea is to use Lowest Common Ancestor of two nodes. There are following cases for R to exist in the path between U and V:
- R is the lowest common ancestor of U and V.
- R is in the left subtree of the lowest common ancestor of U and V and is above V.
- R is in the right subtree of the lowest common ancestor of U and V and is above U.
To know more about the lowest common ancestor, read the post here.
Below is the implementation of the above approach:
C++
// CPP Program to implement the above approach#include <bits/stdc++.h>using namespace std;// Table for storing 2^ith parentvector<vector<int>> table;// Variable to store the height of the treeint height;// Graphvector<vector<int>> Graph;// Arrays to mark start and end time for a nodevector<int> timeIn, timeOut;// Timerint cnt_time;// constructor for initializing// the global variablesvoid initialise(int n){ // log(n) with base 2 height = (int)ceil(log2(n)); // Filling with -1 as initial table.resize(n + 1, vector<int>(height + 1, -1)); // Fill the graph with empty lists Graph.resize(n + 1); timeIn.resize(n + 1); timeOut.resize(n + 1); cnt_time = 0;}// Dfs for pre-processing sparse table and// calculating start and end timevoid dfs(int s, int p){ // Parent at 1 node distance is always // it's direct parent table[s][0] = p; // Start time noted timeIn[s] = ++cnt_time; // Filling sparse table recursively for (int i = 1; i <= height; i++) table[s][i] = table[table[s][i - 1]][i - 1]; // Traversing children of source for (int child : Graph[s]) { if (child == p) continue; dfs(child, s); } // End time noted timeOut[s] = ++cnt_time;}// Helper function to check lowest common Ancestorbool check(int u, int v){ return timeIn[u] <= timeIn[v] && timeOut[u] >= timeOut[v];}// Function to return Lowest Common Ancestor of U and Vint lowestCommonAncestor(int U, int V){ if (check(U, V)) return U; if (check(V, U)) return V; for (int i = height; i >= 0; i--) { if (!check(table[U][i], V)) U = table[U][i]; } return table[U][0];}// Function that return true if R// exists on the path between U// and V in the given treebool isPresent(int U, int V, int R) { // Dfs dfs(1, 1); // Calculating LCA between U and V int LCA = lowestCommonAncestor(U, V); // Calculating LCA between U and R int LCA_1 = lowestCommonAncestor(U, R); // Calculating LCA between U and V int LCA_2 = lowestCommonAncestor(V, R); if (LCA == R || (LCA_1 == LCA && LCA_2 == R) || (LCA_2 == LCA && LCA_1 == R)) { return true; } return false;}// Driver codeint main(int argc, char const *argv[]) { // Number of vertices int n = 6; initialise(n); // Create the graph Graph[1].push_back(2); Graph[2].push_back(1); Graph[1].push_back(3); Graph[3].push_back(1); Graph[2].push_back(4); Graph[4].push_back(2); Graph[2].push_back(5); Graph[5].push_back(2); Graph[3].push_back(6); Graph[6].push_back(3); int U = 4, V = 6, R = 2; if (isPresent(U, V, R)) cout << "Yes" << endl; else cout << "No" << endl;}// This code is contributed by sanjeev2552 |
Java
// Java implementation of the approachimport java.util.*;class GfG { // Table for storing 2^ith parent private static int table[][]; // Variable to store the height of the tree private static int height; // Graph private static ArrayList<ArrayList<Integer> > Graph; // Arrays to mark start and end time for a node private static int timeIn[]; private static int timeOut[]; // Timer private static int time; // Private constructor for initializing // the global variables private GfG(int n) { // log(n) with base 2 height = (int)Math.ceil(Math.log10(n) / Math.log10(2)); table = new int[n + 1][height + 1]; // Fill the graph with empty lists Graph = new ArrayList<ArrayList<Integer> >(); for (int i = 0; i <= n; i++) Graph.add(new ArrayList<Integer>()); timeIn = new int[n + 1]; timeOut = new int[n + 1]; time = 0; } // Filling with -1 as initial private static void preprocessing(int n) { for (int i = 0; i < n + 1; i++) { Arrays.fill(table[i], -1); } } // Dfs for pre-processing sparse table and // calculating start and end time private static void dfs(int s, int p) { // Parent at 1 node distance is always // it's direct parent table[s][0] = p; // Start time noted timeIn[s] = ++time; // Filling sparse table recursively for (int i = 1; i <= height; i++) table[s][i] = table[table[s][i - 1]][i - 1]; // Traversing children of source for (int child : Graph.get(s)) { if (child == p) continue; dfs(child, s); } // End time noted timeOut[s] = ++time; } // Helper function to check lowest common Ancestor private static boolean check(int u, int v) { return timeIn[u] <= timeIn[v] && timeOut[u] >= timeOut[v]; } // Function to return Lowest Common Ancestor of U and V private static int lowestCommonAncestor(int U, int V) { if (check(U, V)) return U; if (check(V, U)) return V; for (int i = height; i >= 0; i--) { if (!check(table[U][i], V)) U = table[U][i]; } return table[U][0]; } // Function that return true if R // exists on the path between U // and V in the given tree private static boolean isPresent(int U, int V, int R) { // Dfs dfs(1, 1); // Calculating LCA between U and V int LCA = lowestCommonAncestor(U, V); // Calculating LCA between U and R int LCA_1 = lowestCommonAncestor(U, R); // Calculating LCA between U and V int LCA_2 = lowestCommonAncestor(V, R); if (LCA == R || (LCA_1 == LCA && LCA_2 == R) || (LCA_2 == LCA && LCA_1 == R)) { return true; } return false; } // Driver code public static void main(String args[]) { // Number of vertices int n = 6; GfG obj = new GfG(n); // Create the graph preprocessing(n); Graph.get(1).add(2); Graph.get(2).add(1); Graph.get(1).add(3); Graph.get(3).add(1); Graph.get(2).add(4); Graph.get(4).add(2); Graph.get(2).add(5); Graph.get(5).add(2); Graph.get(3).add(6); Graph.get(6).add(3); int U = 4, V = 6, R = 2; if (isPresent(U, V, R)) System.out.print("Yes"); else System.out.print("No"); }} |
Python3
# Python3 implementation of the approachimport mathn = 6# GraphGraph = []# log(n) with base 2height = math.ceil(math.log10(n) / math.log10(2))table = [[-1 for i in range(n + 1)] for j in range(n + 1)]# Fill the graph with empty listsfor i in range(n + 1): Graph.append([])timeIn = [0]*(n + 1)timeOut = [0]*(n + 1)time = 0# Filling with -1 as initialdef preprocessing(n): for i in range(n + 1): for j in range(height + 1): table[i][j] = -1# Dfs for pre-processing sparse table and# calculating start and end timedef dfs(s, p): global time # Parent at 1 node distance is always # it's direct parent table[s][0] = p # Start time noted timeIn[s] = time+1 # Filling sparse table recursively for i in range(1, height + 1): table[s][i] = table[table[s][i - 1]][i - 1] # Traversing children of source for child in range(len(Graph[s])): if Graph[s][child] == p: continue dfs(Graph[s][child], s) # End time noted time+=1 timeOut[s] = time# Helper function to check lowest common Ancestordef check(u, v): return (timeIn[u] <= timeIn[v] and timeOut[u] >= timeOut[v])# Function to return Lowest Common Ancestor of U and Vdef lowestCommonAncestor(U, V): if check(U, V): return U if check(V, U): return V for i in range(height, -1, -1): if not check(table[U][i], V): U = table[U][i] return table[U][0]# Function that return true if R# exists on the path between U# and V in the given treedef isPresent(U, V, R): # Dfs dfs(1, 1) # Calculating LCA between U and V LCA = lowestCommonAncestor(U, V) # Calculating LCA between U and R LCA_1 = lowestCommonAncestor(U, R) # Calculating LCA between U and V LCA_2 = lowestCommonAncestor(V, R) if LCA == R or (LCA_1 == LCA and LCA_2 == R) or (LCA_2 == LCA and LCA_1 == R): return True return False# Create the graphpreprocessing(n)Graph[1].append(2)Graph[2].append(1)Graph[1].append(3)Graph[3].append(1)Graph[2].append(4)Graph[4].append(2)Graph[2].append(5)Graph[5].append(2)Graph[3].append(6)Graph[6].append(3)U, V, R = 4, 6, 2if isPresent(U, V, R): print("Yes")else: print("No") # This code is contributed by suresh07. |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;class GfG { // Table for storing 2^ith parent private static int [,]table; // Variable to store the height of the tree private static int height; // Graph private static List<List<int> > Graph; // Arrays to mark start and end time for a node private static int []timeIn; private static int []timeOut; // Timer private static int time; // Private constructor for initializing // the global variables private GfG(int n) { // log(n) with base 2 height = (int)Math.Ceiling(Math.Log10(n) / Math.Log10(2)); table = new int[n + 1, height + 1]; // Fill the graph with empty lists Graph = new List<List<int> >(); for (int i = 0; i <= n; i++) Graph.Add(new List<int>()); timeIn = new int[n + 1]; timeOut = new int[n + 1]; time = 0; } // Filling with -1 as initial private static void preprocessing(int n) { for (int i = 0; i < n + 1; i++) { for(int j = 0; j < height + 1; j++) table[i, j] = -1; } } // Dfs for pre-processing sparse table and // calculating start and end time private static void dfs(int s, int p) { // Parent at 1 node distance is always // it's direct parent table[s, 0] = p; // Start time noted timeIn[s] = ++time; // Filling sparse table recursively for (int i = 1; i <= height; i++) table[s, i] = table[table[s, i - 1], i - 1]; // Traversing children of source foreach (int child in Graph[s]) { if (child == p) continue; dfs(child, s); } // End time noted timeOut[s] = ++time; } // Helper function to check lowest common Ancestor private static bool check(int u, int v) { return timeIn[u] <= timeIn[v] && timeOut[u] >= timeOut[v]; } // Function to return Lowest Common Ancestor of U and V private static int lowestCommonAncestor(int U, int V) { if (check(U, V)) return U; if (check(V, U)) return V; for (int i = height; i >= 0; i--) { if (!check(table[U, i], V)) U = table[U, i]; } return table[U, 0]; } // Function that return true if R // exists on the path between U // and V in the given tree private static bool isPresent(int U, int V, int R) { // Dfs dfs(1, 1); // Calculating LCA between U and V int LCA = lowestCommonAncestor(U, V); // Calculating LCA between U and R int LCA_1 = lowestCommonAncestor(U, R); // Calculating LCA between U and V int LCA_2 = lowestCommonAncestor(V, R); if (LCA == R || (LCA_1 == LCA && LCA_2 == R) || (LCA_2 == LCA && LCA_1 == R)) { return true; } return false; } // Driver code public static void Main(String []args) { // Number of vertices int n = 6; GfG obj = new GfG(n); // Create the graph preprocessing(n); Graph[1].Add(2); Graph[2].Add(1); Graph[1].Add(3); Graph[3].Add(1); Graph[2].Add(4); Graph[4].Add(2); Graph[2].Add(5); Graph[5].Add(2); Graph[3].Add(6); Graph[6].Add(3); int U = 4, V = 6, R = 2; if (isPresent(U, V, R)) Console.Write("Yes"); else Console.Write("No"); }}// This code is contributed by PrinciRaj1992 |
Javascript
<script> // JavaScript implementation of the approach let n = 6; // Table for storing 2^ith parent let table; // Variable to store the height of the tree let height; // Graph let Graph = []; // Arrays to mark start and end time for a node let timeIn; let timeOut; // Timer let time; // log(n) with base 2 height = Math.ceil(Math.log10(n) / Math.log10(2)); table = new Array(n + 1); // Fill the graph with empty lists for (let i = 0; i <= n; i++) Graph.push([]); timeIn = new Array(n + 1); timeOut = new Array(n + 1); time = 0; // Filling with -1 as initial function preprocessing(n) { for (let i = 0; i < n + 1; i++) { table[i] = new Array(height + 1); for(let j = 0; j < height + 1; j++) { table[i][j] = -1; } } } // Dfs for pre-processing sparse table and // calculating start and end time function dfs(s, p) { // Parent at 1 node distance is always // it's direct parent table[s][0] = p; // Start time noted timeIn[s] = ++time; // Filling sparse table recursively for (let i = 1; i <= height; i++) table[s][i] = table[table[s][i - 1]][i - 1]; // Traversing children of source for (let child = 0; child < Graph[s].length; child++) { if (Graph[s][child] == p) continue; dfs(Graph[s][child], s); } // End time noted timeOut[s] = ++time; } // Helper function to check lowest common Ancestor function check(u, v) { return timeIn[u] <= timeIn[v] && timeOut[u] >= timeOut[v]; } // Function to return Lowest Common Ancestor of U and V function lowestCommonAncestor(U, V) { if (check(U, V)) return U; if (check(V, U)) return V; for (let i = height; i >= 0; i--) { if (!check(table[U][i], V)) U = table[U][i]; } return table[U][0]; } // Function that return true if R // exists on the path between U // and V in the given tree function isPresent(U, V, R) { // Dfs dfs(1, 1); // Calculating LCA between U and V let LCA = lowestCommonAncestor(U, V); // Calculating LCA between U and R let LCA_1 = lowestCommonAncestor(U, R); // Calculating LCA between U and V let LCA_2 = lowestCommonAncestor(V, R); if (LCA == R || (LCA_1 == LCA && LCA_2 == R) || (LCA_2 == LCA && LCA_1 == R)) { return true; } return false; } // Create the graph preprocessing(n); Graph[1].push(2); Graph[2].push(1); Graph[1].push(3); Graph[3].push(1); Graph[2].push(4); Graph[4].push(2); Graph[2].push(5); Graph[5].push(2); Graph[3].push(6); Graph[6].push(3); let U = 4, V = 6, R = 2; if (isPresent(U, V, R)) document.write("Yes"); else document.write("No");</script> |
Output:
Yes
Time Complexity: O(NlogN) for pre-processing and logN for finding the lowest common ancestor.
Auxiliary Space: O(NlogN).
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