Print the path between any two nodes of a tree | DFS
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 |
Javascript
<script> // Javascript implementation of the above approach let v = []; // An utility function to add an edge in an // undirected graph. function addEdge(x, y) { v[x].push(y); v[y].push(x); } // A function to print the path between // the given pair of nodes. function printPath(stack) { for (let i = 0; i < stack.length - 1; i++) { document.write(stack[i] + " -> " ); } document.write(stack[stack.length - 1] + "<br>" ); } // An utility function to do // DFS of graph recursively // from a given vertex x. function DFS(vis, x, y, stack) { stack.push(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].length > 0) { for (let j = 0; j < v[x].length; j++) { // If the node is not visited if (vis[v[x][j]] == false ) { DFS(vis, v[x][j], y, stack); } } } stack.pop(); } // A utility function to initialise // visited for the node and call // DFS function for a given vertex x. function DFSCall(x, y, n, stack) { // Visited array let vis = new Array(n + 1); for (let i = 0; i < (n + 1); i++) { vis[i] = false ; } // memset(vis, false, sizeof(vis)) // DFS function call DFS(vis, x, y, stack); } // Driver code for (let i = 0; i < 100; i++) { v.push([]); } let n = 10; let 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 patel2127 </script> |
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; } |
4 -> 3 -> 2 -> 1 -> 7
Time Complexity : O(N)
Space Complexity : O(N)