Added C++ code for lowest common ancestor problem solution using binary lifting

code/graph_algorithms/src/lowest_common_ancestor/lowest_common_ancestor.cpp

@@ -0,0 +1,82 @@
+#include <iostream>
+#include <vector>
+#include <array>
+using namespace std;
+
+// Part of Cosmos by OpenGenus Foundation
+
+// Problem Statement
+/*
+ * Given a tree with n nodes
+ * Your task is to process q queries of the following form:
+ * who is the lowest common ancestor of nodes a and b in the tree?
+ *
+ * The lowest common ancestor of two nodes of a rooted tree is the lowest node
+ * whose subtree contains both the nodes.
+ */
+
+void dfs(int k, int p, vector<int> &depth, vector<vector<int>> &adj, vector<array<int, 20>> &ancestor)
+{
+    ancestor[k][0] = p;
+    depth[k] = depth[p] + 1;
+    for (int i = 1; i < 20; i++)
+        ancestor[k][i] = ancestor[ancestor[k][i - 1]][i - 1];
+
+    for (int i : adj[k])
+        if (i != p)
+            dfs(i, k, depth, adj, ancestor);
+}
+
+int kthAncestor(int k, int x, vector<array<int, 20>> &ancestor)
+{
+    for (int i = 0; i < 20; i++)
+        if (x & (1 << i))
+            k = ancestor[k][i];
+
+    return k;
+}
+
+int lca(int a, int b, vector<array<int, 20>> &ancestor, vector<int> &depth)
+{
+    if (depth[a] < depth[b])
+        swap(a, b);
+
+    a = kthAncestor(a, depth[a] - depth[b], ancestor);
+
+    if (a == b)
+        return b;
+
+    for (int i = 19; i >= 0; i--)
+        if (ancestor[a][i] != ancestor[b][i])
+            a = ancestor[a][i], b = ancestor[b][i];
+
+    return ancestor[a][0];
+}
+
+int main()
+{
+    int n;
+    cin >> n;
+    vector<vector<int>> adj(n + 1);
+    for (int i = 1; i < n; i++)
+    {
+        int a, b;
+        cin >> a >> b;
+        adj[a].push_back(b);
+        adj[b].push_back(a);
+    }
+
+    vector<int> depth(n + 1);
+    vector<array<int, 20>> ancestor(n + 1);
+
+    dfs(1, 0, depth, adj, ancestor);
+
+    int q;
+    cin >> q;
+    while (q--)
+    {
+        int a, b;
+        cin >> a >> b;
+        cout << lca(a, b, ancestor, depth) << '\n';
+    }
+}