# Shortest path between two nodes in array like representation of binary tree

• Difficulty Level : Easy
• Last Updated : 12 Apr, 2022

Consider a binary tree in which each node has two children except the leaf nodes. If a node is labeled as ‘v’ then its right children will be labeled as 2v+1 and left children as 2v. Root is labelled as
Given two nodes labeled as i and j, the task is to find the shortest distance and the path from i to j. And print the path of node i and node j from root node.

Examples:

Input : i = 1, j = 2
Output : 1
Explanation:
Path is 1 2

Input: i = 4, j = 3
Output : 3
Explanation:
Path is 4 2 1 3

This problem is mainly an extension of Find distance between two given keys of a Binary Tree. Here we not only find the shortest distance but also the path.
The distance between the two nodes i and j will be equal to dist(i, LCA(i, j)) + dist(j, LCA(i, j)) where LCA means the lowest common ancestor of nodes labeled as i and j. If a number x is represented in the binary form then 2*x can be represented by appending 0 to the binary representation of x and 2x+1 can be represented by appending 1 to the binary representation of x. This is because when we append 0 all the terms present in the binary form of x shift left, so it gets doubled similarly when we append 1, we get 2x+1. Suppose the binary representation of a node is 1010 this tells us the path of this node from root. First-term ‘1’ represents root second term 0 represents left turn then the third term 1 represents a right turn from the previous node and finally, 0 represents the left turn.

Node 10 in binary form is 1010 and 13 in binary form is 1101 secondly length of the binary representation of any node also tells about its level in a binary tree. Suppose binary representation of i is m length and is [Tex]i_2          [/Tex]…and binary representation of node j is n length [Tex]j_2          [/Tex]……
Thus we know the path of i and j from the root. Find out k such that for all p<=k . This is the LCA of i and j in binary form.So dist(i, LCA(i, j)) will be m – k and dist(j, LCA(i, j)) = n – k. so answer will be m + n – 2k. And printing the path is also not a big issue just store the path of i to LCA and path of j to LCA and concatenate them.

## C++

 `// C++ representation of finding shortest``// distance between node i and j``#include ``using` `namespace` `std;` `// prints the path between node i and node j``void` `ShortestPath(``int` `i, ``int` `j, ``int` `k, ``int` `m, ``int` `n)``{``    ``// path1 stores path of node i to lca and``    ``// path2 stores path of node j to lca``    ``vector<``int``> path1, path2;``    ``int` `x = m - 1;` `    ``// push node i in path1``    ``path1.push_back(i);` `    ``// keep pushing parent of node labelled``    ``// as i to path1 until lca is reached``    ``while` `(x != k) {``        ``path1.push_back(i / 2);``        ``i = i / 2;``        ``x--;``    ``}``    ``int` `y = n - 1;` `    ``// push node j to path2``    ``path2.push_back(j);` `    ``// keep pushing parent of node j till``    ``// lca is reached``    ``while` `(y != k)``    ``{``        ``path2.push_back(j / 2);``        ``j = j / 2;``        ``y--;``    ``}` `    ``// printing path from node i to lca``    ``for` `(``int` `l = 0; l < path1.size(); l++)``        ``cout << path1[l] << ``" "``;` `    ``// printing path from lca to node j``    ``for` `(``int` `l = path2.size() - 2; l >= 0; l--)``        ``cout << path2[l] << ``" "``;``    ``cout << endl;``}` `// returns the shortest distance between``// nodes labelled as i and j``int` `ShortestDistance(``int` `i, ``int` `j)``{``    ``// vector to store binary form of i and j``    ``vector<``int``> v1, v2;` `    ``// finding binary form of i and j``    ``int` `p1 = i;``    ``int` `p2 = j;``    ``while` `(i != 0)``    ``{``        ``v1.push_back(i % 2);``        ``i = i / 2;``    ``}``    ``while` `(j != 0) {``        ``v2.push_back(j % 2);``        ``j = j / 2;``    ``}` `    ``// as binary form will be in reverse order``    ``// reverse the vectors``    ``reverse(v1.begin(), v1.end());``    ``reverse(v2.begin(), v2.end());` `    ``// finding the k that is lca (i, j)``    ``int` `m = v1.size(), n = v2.size(), k = 0;``    ``if` `(m < n)``    ``{``        ``while` `(k < m && v1[k] == v2[k])``            ``k++;``    ``}``    ``else` `{``        ``while` `(k < n && v1[k] == v2[k])``            ``k++;``    ``}` `    ``ShortestPath(p1, p2, k - 1, m, n);``    ``return` `m + n - 2 * k;``}` `// Driver Code``int` `main()``{``    ``cout << ShortestDistance(1, 2) << endl;``    ``cout << ShortestDistance(4, 3) << endl;``    ``return` `0;``}`

## Java

 `// Java representation of finding shortest``// distance between node i and j``import` `java.util.*;` `class` `GFG``{``    ` `// prints the path between node i and node j``static` `void` `ShortestPath(``int` `i, ``int` `j, ``int` `k, ``int` `m,``                                    ``int` `n)``{``    ``// path1 stores path of node i to lca and``    ``// path2 stores path of node j to lca``    ``Vector path1=``new` `Vector(),``                    ``path2=``new` `Vector();``    ``int` `x = m - ``1``;` `    ``// push node i in path1``    ``path1.add(i);` `    ``// keep pushing parent of node labelled``    ``// as i to path1 until lca is reached``    ``while` `(x != k)``    ``{``        ``path1.add(i / ``2``);``        ``i = i / ``2``;``        ``x--;``    ``}``    ``int` `y = n - ``1``;` `    ``// push node j to path2``    ``path2.add(j);` `    ``// keep pushing parent of node j till``    ``// lca is reached``    ``while` `(y != k)``    ``{``        ``path2.add(j / ``2``);``        ``j = j / ``2``;``        ``y--;``    ``}` `    ``// printing path from node i to lca``    ``for` `(``int` `l = ``0``; l < path1.size(); l++)``        ``System.out.print( path1.get(l) + ``" "``);` `    ``// printing path from lca to node j``    ``for` `(``int` `l = path2.size() - ``2``; l >= ``0``; l--)``        ``System.out.print( path2.get(l) + ``" "``);``    ``System.out.println();``}` `// returns the shortest distance between``// nodes labelled as i and j``static` `int` `ShortestDistance(``int` `i, ``int` `j)``{``    ``// vector to store binary form of i and j``    ``Vector v1=``new` `Vector(),``                    ``v2=``new` `Vector();` `    ``// finding binary form of i and j``    ``int` `p1 = i;``    ``int` `p2 = j;``    ``while` `(i != ``0``)``    ``{``        ``v1.add(i % ``2``);``        ``i = i / ``2``;``    ``}``    ``while` `(j != ``0``)``    ``{``        ``v2.add(j % ``2``);``        ``j = j / ``2``;``    ``}` `    ``// as binary form will be in reverse order``    ``// reverse the vectors``    ``Collections.reverse(v1);``    ``Collections.reverse(v2);` `    ``// finding the k that is lca (i, j)``    ``int` `m = v1.size(), n = v2.size(), k = ``0``;``    ``if` `(m < n)``    ``{``        ``while` `(k < m && v1.get(k) == v2.get(k))``            ``k++;``    ``}``    ``else``    ``{``        ``while` `(k < n && v1.get(k) == v2.get(k))``            ``k++;``    ``}` `    ``ShortestPath(p1, p2, k - ``1``, m, n);``    ``return` `m + n - ``2` `* k;``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``System.out.println( ShortestDistance(``1``, ``2``) );``    ``System.out.println(ShortestDistance(``4``, ``3``) );``}``}` `// This code is contributed by Arnab Kundu`

## Python3

 `# Python3 representation of finding``# shortest distance between node i and j` `# Prints the path between node i and node j``def` `ShortestPath(i, j, k, m, n):` `    ``# path1 stores path of node i to lca and``    ``# path2 stores path of node j to lca``    ``path1, path2 ``=` `[], []``    ``x ``=` `m ``-` `1` `    ``# push node i in path1``    ``path1.append(i)` `    ``# keep pushing parent of node labelled``    ``# as i to path1 until lca is reached``    ``while` `x !``=` `k:``        ``path1.append(i ``/``/` `2``)``        ``i ``=` `i ``/``/` `2``        ``x ``-``=` `1``    ` `    ``y ``=` `n ``-` `1` `    ``# push node j to path2``    ``path2.append(j)` `    ``# keep pushing parent of node``    ``# j till lca is reached``    ``while` `y !``=` `k:``        ``path2.append(j ``/` `2``)``        ``j ``=` `j ``/``/` `2``        ``y ``-``=` `1``    ` `    ``# printing path from node i to lca``    ``for` `l ``in` `range``(``0``, ``len``(path1)):``        ``print``(path1[l], end``=``" "``)` `    ``# printing path from lca to node j``    ``for` `l ``in` `range``(``len``(path2) ``-` `2``, ``-``1``, ``-``1``):``        ``print``(path2[l], end``=``" "``)``    ``print``()` `# Returns the shortest distance``# between nodes labelled as i and j``def` `ShortestDistance(i, j):` `    ``# vector to store binary form of i and j``    ``v1, v2 ``=` `[], []` `    ``# finding binary form of i and j``    ``p1, p2 ``=` `i, j``    ``while` `i !``=` `0``:``        ``v1.append(i ``%` `2``)``        ``i ``=` `i ``/``/` `2``    ` `    ``while` `j !``=` `0``:``        ``v2.append(j ``%` `2``)``        ``j ``=` `j ``/``/` `2``    ` `    ``# as binary form will be in reverse``    ``# order reverse the vectors``    ``v1 ``=` `v1[::``-``1``]``    ``v2 ``=` `v2[::``-``1``]` `    ``# finding the k that is lca (i, j)``    ``m, n, k ``=` `len``(v1), ``len``(v2), ``0``    ``if` `m < n:``        ``while` `k < m ``and` `v1[k] ``=``=` `v2[k]:``            ``k ``+``=` `1``    ` `    ``else``:``        ``while` `k < n ``and` `v1[k] ``=``=` `v2[k]:``            ``k ``+``=` `1``    ` `    ``ShortestPath(p1, p2, k ``-` `1``, m, n)``    ``return` `m ``+` `n ``-` `2` `*` `k` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``print``(ShortestDistance(``1``, ``2``))``    ``print``(ShortestDistance(``4``, ``3``))` `# This code is contributed by Rituraj Jain`

## C#

 `// C#  representation of finding shortest``// distance between node i and j``using` `System;``using` `System.Collections.Generic;   ``    ` `class` `GFG``{``     ` `// prints the path between node i and node j``static` `void` `ShortestPath(``int` `i, ``int` `j, ``int` `k, ``int` `m,``                                    ``int` `n)``{``    ``// path1 stores path of node i to lca and``    ``// path2 stores path of node j to lca``    ``List<``int``> path1=``new` `List<``int``>(),``                    ``path2=``new` `List<``int``>();``    ``int` `x = m - 1;`` ` `    ``// push node i in path1``    ``path1.Add(i);`` ` `    ``// keep pushing parent of node labelled``    ``// as i to path1 until lca is reached``    ``while` `(x != k)``    ``{``        ``path1.Add(i / 2);``        ``i = i / 2;``        ``x--;``    ``}``    ``int` `y = n - 1;`` ` `    ``// push node j to path2``    ``path2.Add(j);`` ` `    ``// keep pushing parent of node j till``    ``// lca is reached``    ``while` `(y != k)``    ``{``        ``path2.Add(j / 2);``        ``j = j / 2;``        ``y--;``    ``}`` ` `    ``// printing path from node i to lca``    ``for` `(``int` `l = 0; l < path1.Count; l++)``        ``Console.Write( path1[l] + ``" "``);`` ` `    ``// printing path from lca to node j``    ``for` `(``int` `l = path2.Count - 2; l >= 0; l--)``        ``Console.Write( path2[l] + ``" "``);``    ``Console.WriteLine();``}`` ` `// returns the shortest distance between``// nodes labelled as i and j``static` `int` `ShortestDistance(``int` `i, ``int` `j)``{``    ``// vector to store binary form of i and j``    ``List<``int``> v1=``new` `List<``int``>(),``                    ``v2=``new` `List<``int``>();`` ` `    ``// finding binary form of i and j``    ``int` `p1 = i;``    ``int` `p2 = j;``    ``while` `(i != 0)``    ``{``        ``v1.Add(i % 2);``        ``i = i / 2;``    ``}``    ``while` `(j != 0)``    ``{``        ``v2.Add(j % 2);``        ``j = j / 2;``    ``}`` ` `    ``// as binary form will be in reverse order``    ``// reverse the vectors``    ``v1.Reverse();``    ``v2.Reverse();`` ` `    ``// finding the k that is lca (i, j)``    ``int` `m =v1.Count, n =v2.Count, k = 0;``    ``if` `(m < n)``    ``{``        ``while` `(k < m && v1[k] == v2[k])``            ``k++;``    ``}``    ``else``    ``{``        ``while` `(k < n && v1[k] == v2[k])``            ``k++;``    ``}`` ` `    ``ShortestPath(p1, p2, k - 1, m, n);``    ``return` `m + n - 2 * k;``}`` ` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``Console.WriteLine( ShortestDistance(1, 2) );``    ``Console.WriteLine(ShortestDistance(4, 3) );``}``}` `// This code is contributed by Princi Singh`

## Javascript

 ``

Output:

```1 2
1
4 2 1 3
3```

Complexity Analysis:
Time Complexity: O(i + j) since we are shifting value of i and j to the right for every traversal

Auxiliary Space: O(log_2 i + log_2 j) since we are storing paths for every right shifting values of i and j.

This article is contributed by Ayush Jha. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.