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Print the path between any two nodes of a tree | DFS
  • Difficulty Level : Medium
  • Last Updated : 11 May, 2021

Given a tree of distinct nodes N with N-1 edges and a pair of nodes P. The task is to find and print the path between the two given nodes of the tree using DFS.
 

Input: N = 10
          1
       /    \
      2      3
    / | \  / | \
   4  5 6  7 8  9
Pair = {4, 8}
Output: 4 -> 2 -> 1 -> 3 -> 8

Input: N = 3
      1
     /  \
    2    3
Pair = {2, 3}
Output:  2 -> 1 -> 3

 

For example, in the above tree the path between nodes 5 and 3 is 5 -> 2 -> 1 -> 3
Path between nodes 4 and 8 is 4 -> 2 -> 1 -> 3 -> 8.
 

Approach: 



  • The idea is to run DFS from the source node and push the traversed nodes into a stack till the destination node is traversed.
  • Whenever backtracking occurs pop the node from the stack.

Note: There should be a path between the given pair of nodes.
Below is the implementation of the above approach: 
 

C++




// C++ implementation
#include <bits/stdc++.h>
using namespace std;
 
// An utility function to add an edge in an
// undirected graph.
void addEdge(vector<int> v[],
             int x,
             int y)
{
    v[x].push_back(y);
    v[y].push_back(x);
}
 
// A function to print the path between
// the given pair of nodes.
void printPath(vector<int> stack)
{
    int i;
    for (i = 0; i < (int)stack.size() - 1;
         i++) {
        cout << stack[i] << " -> ";
    }
    cout << stack[i];
}
 
// An utility function to do
// DFS of graph recursively
// from a given vertex x.
void DFS(vector<int> v[],
         bool vis[],
         int x,
         int y,
         vector<int> stack)
{
    stack.push_back(x);
    if (x == y) {
 
        // print the path and return on
        // reaching the destination node
        printPath(stack);
        return;
    }
    vis[x] = true;
 
    // if backtracking is taking place
    if (!v[x].empty()) {
        for (int j = 0; j < v[x].size(); j++) {
            // if the node is not visited
            if (vis[v[x][j]] == false)
                DFS(v, vis, v[x][j], y, stack);
        }
    }
 
    stack.pop_back();
}
 
// A utility function to initialise
// visited for the node and call
// DFS function for a given vertex x.
void DFSCall(int x,
             int y,
             vector<int> v[],
             int n,
             vector<int> stack)
{
    // visited array
    bool vis[n + 1];
 
    memset(vis, false, sizeof(vis));
 
    // DFS function call
    DFS(v, vis, x, y, stack);
}
 
// Driver Code
int main()
{
    int n = 10;
    vector<int> v[n], stack;
 
    // Vertex numbers should be from 1 to 9.
    addEdge(v, 1, 2);
    addEdge(v, 1, 3);
    addEdge(v, 2, 4);
    addEdge(v, 2, 5);
    addEdge(v, 2, 6);
    addEdge(v, 3, 7);
    addEdge(v, 3, 8);
    addEdge(v, 3, 9);
 
    // Function Call
    DFSCall(4, 8, v, n, stack);
 
    return 0;
}

Java




// Java implementation of the above approach
import java.util.*;
class GFG
{
    static Vector<Vector<Integer>> v = new Vector<Vector<Integer>>();
     
    // An utility function to add an edge in an
    // undirected graph.
    static void addEdge(int x, int y){
        v.get(x).add(y);
        v.get(y).add(x);
    }
     
    // A function to print the path between
    // the given pair of nodes.
    static void printPath(Vector<Integer> stack)
    {
        for(int i = 0; i < stack.size() - 1; i++)
        {
            System.out.print(stack.get(i) + " -> ");
        }
        System.out.println(stack.get(stack.size() - 1));
    }
     
    // An utility function to do
    // DFS of graph recursively
    // from a given vertex x.
    static void DFS(boolean vis[], int x, int y, Vector<Integer> stack)
    {
        stack.add(x);
        if (x == y)
        {
           
            // print the path and return on
            // reaching the destination node
            printPath(stack);
            return;
        }
        vis[x] = true;
       
        // if backtracking is taking place     
        if (v.get(x).size() > 0)
        {
            for(int j = 0; j < v.get(x).size(); j++)
            {
               
                // if the node is not visited
                if (vis[v.get(x).get(j)] == false)
                {
                    DFS(vis, v.get(x).get(j), y, stack);
                }
            }
        }
         
        stack.remove(stack.size() - 1);
    }
     
    // A utility function to initialise
    // visited for the node and call
    // DFS function for a given vertex x.
    static void DFSCall(int x, int y, int n,
                        Vector<Integer> stack)
    {
       
        // visited array
        boolean vis[] = new boolean[n + 1];
        Arrays.fill(vis, false);
       
        // memset(vis, false, sizeof(vis))
       
        // DFS function call
        DFS(vis, x, y, stack);
    }
     
  // Driver code
    public static void main(String[] args)
    {
        for(int i = 0; i < 100; i++)
        {
            v.add(new Vector<Integer>());
        }
     
        int n = 10;
        Vector<Integer> stack = new Vector<Integer>();
           
        // Vertex numbers should be from 1 to 9.
        addEdge(1, 2);
        addEdge(1, 3);
        addEdge(2, 4);
        addEdge(2, 5);
        addEdge(2, 6);
        addEdge(3, 7);
        addEdge(3, 8);
        addEdge(3, 9);
           
        // Function Call
        DFSCall(4, 8, n, stack);
    }
}
 
// This code is contributed by divyeshrabadiya07

Python3




# Python3 implementation of the above approach
v = [[] for i in range(100)]
 
# An utility function to add an edge in an
# undirected graph.
def addEdge(x, y):
    v[x].append(y)
    v[y].append(x)
 
# A function to print the path between
# the given pair of nodes.
def printPath(stack):
    for i in range(len(stack) - 1):
        print(stack[i], end = " -> ")
    print(stack[-1])
 
# An utility function to do
# DFS of graph recursively
# from a given vertex x.
def DFS(vis, x, y, stack):
    stack.append(x)
    if (x == y):
 
        # print the path and return on
        # reaching the destination node
        printPath(stack)
        return
    vis[x] = True
 
    # if backtracking is taking place
 
    if (len(v[x]) > 0):
        for j in v[x]:
             
            # if the node is not visited
            if (vis[j] == False):
                DFS(vis, j, y, stack)
                 
    del stack[-1]
 
# A utility function to initialise
# visited for the node and call
# DFS function for a given vertex x.
def DFSCall(x, y, n, stack):
     
    # visited array
    vis = [0 for i in range(n + 1)]
 
    #memset(vis, false, sizeof(vis))
 
    # DFS function call
    DFS(vis, x, y, stack)
 
# Driver Code
n = 10
stack = []
 
# Vertex numbers should be from 1 to 9.
addEdge(1, 2)
addEdge(1, 3)
addEdge(2, 4)
addEdge(2, 5)
addEdge(2, 6)
addEdge(3, 7)
addEdge(3, 8)
addEdge(3, 9)
 
# Function Call
DFSCall(4, 8, n, stack)
     
# This code is contributed by Mohit Kumar

C#




// C# implementation of the above approach
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG
{
    static List<List<int>> v = new List<List<int>>();
     
    // An utility function to Add an edge in an
    // undirected graph.
    static void addEdge(int x, int y)
    {
        v[x].Add(y);
        v[y].Add(x);
    }
     
    // A function to print the path between
    // the given pair of nodes.
    static void printPath(List<int> stack)
    {
        for(int i = 0; i < stack.Count - 1; i++)
        {
            Console.Write(stack[i] + " -> ");
        }
        Console.WriteLine(stack[stack.Count - 1]);
    }
     
    // An utility function to do
    // DFS of graph recursively
    // from a given vertex x.
    static void DFS(bool []vis, int x, int y, List<int> stack)
    {
        stack.Add(x);
        if (x == y)
        {
           
            // print the path and return on
            // reaching the destination node
            printPath(stack);
            return;
        }
        vis[x] = true;
       
        // if backtracking is taking place     
        if (v[x].Count > 0)
        {
            for(int j = 0; j < v[x].Count; j++)
            {
               
                // if the node is not visited
                if (vis[v[x][j]] == false)
                {
                    DFS(vis, v[x][j], y, stack);
                }
            }
        }        
        stack.RemoveAt(stack.Count - 1);
    }
     
    // A utility function to initialise
    // visited for the node and call
    // DFS function for a given vertex x.
    static void DFSCall(int x, int y, int n,
                        List<int> stack)
    {
       
        // visited array
        bool []vis = new bool[n + 1];
        Array.Fill(vis, false);
       
        // memset(vis, false, sizeof(vis))
       
        // DFS function call
        DFS(vis, x, y, stack);
    }
     
  // Driver code
    public static void Main(string[] args)
    {
        for(int i = 0; i < 100; i++)
        {
            v.Add(new List<int>());
        }
     
        int n = 10;
        List<int> stack = new List<int>();
           
        // Vertex numbers should be from 1 to 9.
        addEdge(1, 2);
        addEdge(1, 3);
        addEdge(2, 4);
        addEdge(2, 5);
        addEdge(2, 6);
        addEdge(3, 7);
        addEdge(3, 8);
        addEdge(3, 9);
           
        // Function Call
        DFSCall(4, 8, n, stack);
    }
}
 
// This code is contributed by rutvik_56
Output: 
4 -> 2 -> 1 -> 3 -> 8

 

Efficient Approach : 

In this approach we will utilise the concept of Lowest Common Ancestor (LCA).

1. We will find level and parent of every node using DFS.

2. We will find lowest common ancestor (LCA) of the two given nodes.

3. Starting from the first node we will travel to the LCA and keep on pushing

the intermediates nodes in our path vector.



4. Then, from the second node we will again travel to the LCA but this time

we will reverse the encountered intermediate nodes and then push them in

our path vector.

5. Finally, print the path vector to get the path between the two nodes.

C++




// C++ implementation for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// An utility function to add an edge in the tree
void addEdge(vector<int> adj[], int x,
             int y)
{
    adj[x].push_back(y);
    adj[y].push_back(x);
}
 
// running dfs to find level and parent of every node
void dfs(vector<int> adj[], int node, int l,
            int p, int lvl[], int par[])
{
   lvl[node] = l;
   par[node] = p;
   
   for(int child : adj[node])
   {
      if(child != p)
        dfs(adj, child, l+1, node, lvl, par);
   }
}
 
int LCA(int a, int b, int par[], int lvl[])
   // if node a is at deeper level than
   // node b
   if(lvl[a] > lvl[b])
    swap(a, b);
    
   // finding the difference in levels
   // of node a and b
   int diff = lvl[b] - lvl[a];
    
   // moving b to the level of a
   while(diff != 0)
   {
      b = par[b];
      diff--;
   }
    
   // means we have found the LCA
   if(a == b)
    return a;
    
   // finding the LCA
   while(a != b)
    a=par[a], b=par[b];
 
   return a;
}
 
void printPath(vector<int> adj[], int a, int b, int n)
{
    // stores level of every node
    int lvl[n+1];
   
    // stores parent of every node
    int par[n+1];
   
    // running dfs to find parent and level
    // of every node in the tree
    dfs(adj, 1, 0, -1, lvl, par);
     
    // finding the lowest common ancestor
    // of the nodes a and b
    int lca = LCA(a, b, par, lvl);
   
    // stores path between nodes a and b
    vector<int> path;
     
    // traversing the path from a to lca
    while(a != lca)
      path.push_back(a), a = par[a];
 
    path.push_back(a);
 
    vector<int> temp;
     
    // traversing the path from b to lca
    while(b != lca)
      temp.push_back(b), b=par[b];
     
    // reversing the path to get actual path
    reverse(temp.begin(), temp.end());
   
    for(int x : temp)
      path.push_back(x);
   
    // printing the path
    for(int i = 0; i < path.size() - 1; i++)
      cout << path[i] << " -> ";
   
    cout << path[path.size() - 1] << endl;
}
 
// Driver Code
int main()
{  
  /*                          1
  
                        /            \
 
                     2                7
 
               /             \
 
             3                6
 
    /        |        \
 
  4          8          5
   
 */
    // number of nodes in the tree
    int n = 8;
     
    // adjacency list representation of the tree
    vector<int> adj[n+1];
   
    addEdge(adj, 1, 2);
    addEdge(adj, 1, 7);
    addEdge(adj, 2, 3);
    addEdge(adj, 2, 6);
    addEdge(adj, 3, 4);
    addEdge(adj, 3, 8);
    addEdge(adj, 3, 5);
     
    // taking two input nodes
    // between which path is
    // to be printed
    int a = 4, b = 7;
   
    printPath(adj, a, b, n);
    
    return 0;
}
Output
4 -> 3 -> 2 -> 1 -> 7

Time Complexity : O(N) 

Space Complexity : O(N)

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