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 Path is 1 2 Input : i = 4, j = 3 Output : 3 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 shortest distance, but also the path.

The between the two nodes i and j will be equal to dist(i, LCA(i, j)) + dist(j, LCA(i, j)) where LCA means lowest common ancestor of nodes labeled as i and j. If a number x is represented in 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 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 third term 1 represents right turn from previous node and finally 0 represents left turn.

Node 10 in binary form is 1010 and 13 in binary form is 1101 secondly length of binary representation of any node also tells about its level in binary tree. Suppose binary representation of i is m length and is … and binary representation of node j is n length …….

Thus we know the path of i and j from 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++ representation of finding shortest // distance between node i and j #include <bits/stdc++.h> 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 function int main() { cout << ShortestDistance(1, 2) << endl; cout << ShortestDistance(4, 3) << endl; return 0; }

Output:

1 2 1 4 2 1 3 3

**Time Complexity** O(i + j)

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

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