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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 ``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> v = ``new` `Vector>();``    ` `    ``// 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 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 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 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());``        ``}``    ` `        ``int` `n = ``10``;``        ``Vector stack = ``new` `Vector();``          ` `        ``// 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> v = ``new` `List>();``    ` `    ``// 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 ``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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