Skip to content
Related Articles

Related Articles

Check if two nodes are on same path in a tree | Set 2
  • Difficulty Level : Hard
  • Last Updated : 25 Aug, 2020

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Output: 

Yes

Time Complexity: O(N) 
Auxiliary Space: O(1)
 

competitive-programming-img




My Personal Notes arrow_drop_up
Recommended Articles
Page :