Check if two nodes are on same path in a tree | Set 2

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Given two nodes of a binary tree v1 and v2, the task is to check if two nodes are on the same path in a tree.
Example:

Input:  v1 = 1, v2 = 5
       1
    /  |  \
   2   3   4
  /    |    \
 5     6     7
 
Output: Yes
Explanation:
Both nodes 1 and 5
lie in the path 1 -> 2 -> 5.

Input: v1 = 2, v2 = 6
       1
    /  |  \
   2   3   4
  /    |    \
 5     6     7

Output: NO

DFS Approach: Refer to Check if two nodes are on same path in a tree for the DFS approach.
LCA Approach: The idea is to use Lowest Common Ancestor. Find the LCA of the given vertices v1 and v2. If the LCA is equal to any of the given two vertices, print Yes. Otherwise, print No.
Below is the implementation of above approach:

C++

// C++ program to check if two nodes
// are on same path in a tree without
// using any extra space
#include <bits/stdc++.h>
using namespace std;
 
// Function to filter
// the return Values
int filter(int x, int y, int z)
{
    if (x != -1 && y != -1) {
        return z;
    }
    return x == -1 ? y : x;
}
 
// Utility function to check if nodes
// are on same path or not
int samePathUtil(int mtrx[][7], int vrtx,
                 int v1, int v2, int i)
{
    int ans = -1;
 
    // Condition to check
    // if any vertex
    // is equal to given two
    // vertex or not
    if (i == v1 || i == v2)
        return i;
 
    for (int j = 0; j < vrtx; j++) {
 
        // Check if the current
        // position has 1
        if (mtrx[i][j] == 1) {
            // Recursive call
            ans
                = filter(ans,
                         samePathUtil(mtrx,
                                      vrtx,
                                      v1,
                                      v2,
                                      j),
                         i);
        }
    }
 
    // Return LCA
    return ans;
}
 
// Function to check if nodes
// lies on same path or not
bool isVertexAtSamePath(int mtrx[][7],
                        int vrtx, int v1,
                        int v2, int i)
{
    int lca = samePathUtil(mtrx,
                           vrtx, v1 - 1,
                           v2 - 1, i);
 
    if (lca == v1 - 1 || lca == v2 - 1)
        return true;
 
    return false;
}
 
// Driver Program
int main()
{
    int vrtx = 7, edge = 6;
    int mtrx[7][7] = {
        { 0, 1, 1, 1, 0, 0, 0 },
        { 0, 0, 0, 0, 1, 0, 0 },
        { 0, 0, 0, 0, 0, 1, 0 },
        { 0, 0, 0, 0, 0, 0, 1 },
        { 0, 0, 0, 0, 0, 0, 0 },
        { 0, 0, 0, 0, 0, 0, 0 },
        { 0, 0, 0, 0, 0, 0, 0 }
 
    };
 
    int v1 = 1, v2 = 5;
 
    if (isVertexAtSamePath(mtrx,
                           vrtx, v1,
                           v2, 0))
        cout << "Yes";
 
    else
        cout << "No";
 
    return 0;
}

Java

// Java program to check if two nodes
// are on same path in a tree without
// using any extra space
class GFG{
 
// Function to filter
// the return Values
static int filter(int x, int y, int z)
{
    if (x != -1 && y != -1)
    {
        return z;
    }
    return x == -1 ? y : x;
}
 
// Utility function to check if nodes
// are on same path or not
static int samePathUtil(int mtrx[][], int vrtx,
                        int v1, int v2, int i)
{
    int ans = -1;
 
    // Condition to check
    // if any vertex
    // is equal to given two
    // vertex or not
    if (i == v1 || i == v2)
        return i;
 
    for(int j = 0; j < vrtx; j++)
    {
         
        // Check if the current
        // position has 1
        if (mtrx[i][j] == 1)
        {
             
            // Recursive call
            ans = filter(ans, samePathUtil(
                         mtrx, vrtx, v1,
                         v2, j), i);
        }
    }
 
    // Return LCA
    return ans;
}
 
// Function to check if nodes
// lies on same path or not
static boolean isVertexAtSamePath(int mtrx[][],
                                  int vrtx, int v1,
                                  int v2, int i)
{
    int lca = samePathUtil(mtrx, vrtx, v1 - 1,
                                       v2 - 1, i);
                                        
    if (lca == v1 - 1 || lca == v2 - 1)
        return true;
 
    return false;
}
 
// Driver code
public static void main(String[] args)
{
    int vrtx = 7;
    int mtrx[][] = { { 0, 1, 1, 1, 0, 0, 0 },
                     { 0, 0, 0, 0, 1, 0, 0 },
                     { 0, 0, 0, 0, 0, 1, 0 },
                     { 0, 0, 0, 0, 0, 0, 1 },
                     { 0, 0, 0, 0, 0, 0, 0 },
                     { 0, 0, 0, 0, 0, 0, 0 },
                     { 0, 0, 0, 0, 0, 0, 0 } };
 
    int v1 = 1, v2 = 5;
 
    if (isVertexAtSamePath(mtrx, vrtx,
                           v1, v2, 0))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program to check if two nodes
# are on same path in a tree without
# using any extra space
 
# Function to filter
# the return Values
def filter(x, y, z):
     
    if (x != -1 and y != -1):
        return z
     
    return y if x == -1 else x
 
# Utility function to check if nodes
# are on same path or not
def samePathUtil(mtrx, vrtx, v1, v2, i):
     
    ans = -1
 
    # Condition to check
    # if any vertex
    # is equal to given two
    # vertex or not
    if (i == v1 or i == v2):
        return i
     
    for j in range(0, vrtx):
         
        # Check if the current
        # position has 1
        if (mtrx[i][j] == 1):
             
            # Recursive call
            ans = filter(ans, 
                         samePathUtil(mtrx, vrtx, 
                                   v1, v2, j), i)
         
    # Return LCA
    return ans
 
# Function to check if nodes
# lies on same path or not
def isVertexAtSamePath(mtrx, vrtx, v1, v2, i):
     
    lca = samePathUtil(mtrx, vrtx, v1 - 1,
                                   v2 - 1, i)
 
    if (lca == v1 - 1 or lca == v2 - 1):
        return True
 
    return False
 
# Driver code
vrtx = 7
edge = 6
 
mtrx = [ [ 0, 1, 1, 1, 0, 0, 0 ] ,
         [ 0, 0, 0, 0, 1, 0, 0 ],
         [ 0, 0, 0, 0, 0, 1, 0 ],
         [ 0, 0, 0, 0, 0, 0, 1 ],
         [ 0, 0, 0, 0, 0, 0, 0 ],
         [ 0, 0, 0, 0, 0, 0, 0 ],
         [ 0, 0, 0, 0, 0, 0, 0 ] ]
 
v1 = 1
v2 = 5
 
if (isVertexAtSamePath(mtrx, vrtx, v1, v2, 0)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by sanjoy_62

C#

// C# program to check if two nodes
// are on same path in a tree without
// using any extra space
using System;
class GFG{
 
// Function to filter
// the return Values
static int filter(int x, int y, int z)
{
    if (x != -1 && y != -1)
    {
        return z;
    }
    return x == -1 ? y : x;
}
 
// Utility function to check if nodes
// are on same path or not
static int samePathUtil(int [,]mtrx, int vrtx,
                        int v1, int v2, int i)
{
    int ans = -1;
 
    // Condition to check
    // if any vertex
    // is equal to given two
    // vertex or not
    if (i == v1 || i == v2)
        return i;
 
    for(int j = 0; j < vrtx; j++)
    {
         
        // Check if the current
        // position has 1
        if (mtrx[i,j] == 1)
        {
             
            // Recursive call
            ans = filter(ans, samePathUtil(
                         mtrx, vrtx, v1,
                         v2, j), i);
        }
    }
 
    // Return LCA
    return ans;
}
 
// Function to check if nodes
// lies on same path or not
static bool isVertexAtSamePath(int [,]mtrx,
                               int vrtx, int v1,
                               int v2, int i)
{
    int lca = samePathUtil(mtrx, vrtx, v1 - 1,
                                       v2 - 1, i);
                                        
    if (lca == v1 - 1 || lca == v2 - 1)
        return true;
 
    return false;
}
 
// Driver code
public static void Main(String[] args)
{
    int vrtx = 7;
    int [,]mtrx = { { 0, 1, 1, 1, 0, 0, 0 },
                    { 0, 0, 0, 0, 1, 0, 0 },
                    { 0, 0, 0, 0, 0, 1, 0 },
                    { 0, 0, 0, 0, 0, 0, 1 },
                    { 0, 0, 0, 0, 0, 0, 0 },
                    { 0, 0, 0, 0, 0, 0, 0 },
                    { 0, 0, 0, 0, 0, 0, 0 } };
 
    int v1 = 1, v2 = 5;
 
    if (isVertexAtSamePath(mtrx, vrtx,
                           v1, v2, 0))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by sapnasingh4991

Javascript

<script>
    // Javascript program to check if two nodes
    // are on same path in a tree without
    // using any extra space
     
    // Function to filter
    // the return Values
    function filter(x, y, z)
    {
        if (x != -1 && y != -1)
        {
            return z;
        }
        return x == -1 ? y : x;
    }
 
    // Utility function to check if nodes
    // are on same path or not
    function samePathUtil(mtrx, vrtx, v1, v2, i)
    {
        let ans = -1;
 
        // Condition to check
        // if any vertex
        // is equal to given two
        // vertex or not
        if (i == v1 || i == v2)
            return i;
 
        for(let j = 0; j < vrtx; j++)
        {
 
            // Check if the current
            // position has 1
            if (mtrx[i][j] == 1)
            {
 
                // Recursive call
                ans = filter(ans, samePathUtil(
                             mtrx, vrtx, v1,
                             v2, j), i);
            }
        }
 
        // Return LCA
        return ans;
    }
 
    // Function to check if nodes
    // lies on same path or not
    function isVertexAtSamePath(mtrx, vrtx, v1, v2, i)
    {
        let lca = samePathUtil(mtrx, vrtx, v1 - 1, v2 - 1, i);
 
        if (lca == v1 - 1 || lca == v2 - 1)
            return true;
 
        return false;
    }
     
    let vrtx = 7;
    let mtrx = [ [ 0, 1, 1, 1, 0, 0, 0 ],
                  [ 0, 0, 0, 0, 1, 0, 0 ],
                  [ 0, 0, 0, 0, 0, 1, 0 ],
                  [ 0, 0, 0, 0, 0, 0, 1 ],
                  [ 0, 0, 0, 0, 0, 0, 0 ],
                  [ 0, 0, 0, 0, 0, 0, 0 ],
                  [ 0, 0, 0, 0, 0, 0, 0 ] ];
   
    let v1 = 1, v2 = 5;
   
    if (isVertexAtSamePath(mtrx, vrtx,
                           v1, v2, 0))
        document.write("Yes");
    else
        document.write("No");
     
// This code is contributed by mukesh07.    
</script>

Output:

Yes

Time Complexity: O(N²) where N is the number of vertices in the graph.
Auxiliary Space: O(1)

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