Given a rooted tree with N vertices and N-1 edges. We will be given many pairs of vertices u and v, we need to tell whether u is an ancestor of v or not. Given tree will be rooted at the vertex with index 0.

Examples:

u = 1 v = 6 we can see from above tree that node 1 is ancestor of node 6 so the answer will be yes. u = 1 v = 7 we can see from above tree that node 1 is not an ancestor of node 7 so the answer will be no.

We can solve this problem using depth first search of the tree. While doing dfs we can observe a relation between the order in which we visit a node and its ancestors. If we assign in-time and out-time to each node when entering and leaving that node in dfs then we can see that for each pair of ancestor-descendant the in-time of ancestor is less than that of descendant and out-time of ancestor is more than that of descendant, so using this relation we can find the result for each pair of node in O(1) time.

So time complexity for preprocessing will be O(N) and for the query it will be O(1).

// C/C++ program to query whether two node has // ancestor-descendant relationship or not #include <bits/stdc++.h> using namespace std; // Utility dfs method to assign in and out time // to each node void dfs(vector<int> g[], int u, int parent, int timeIn[], int timeOut[], int& cnt) { // assign In-time to node u timeIn[u] = cnt++; // call dfs over all neighbors except parent for (int i = 0; i < g[u].size(); i++) { int v = g[u][i]; if (v != parent) dfs(g, v, u, timeIn, timeOut, cnt); } // assign Out-time to node u timeOut[u] = cnt++; } // method to preprocess all nodes for assigning time void preProcess(int edges[][2], int V, int timeIn[], int timeOut[]) { vector<int> g[V]; // construct array of vector data structure // for tree for (int i = 0; i < V - 1; i++) { int u = edges[i][0]; int v = edges[i][1]; g[u].push_back(v); g[v].push_back(u); } int cnt = 0; // call dfs method from root dfs(g, 0, -1, timeIn, timeOut, cnt); } // method returns "yes" if u is a ancestor // node of v string isAncestor(int u, int v, int timeIn[], int timeOut[]) { bool b = (timeIn[u] <= timeIn[v] && timeOut[v] <= timeOut[u]); return (b ? "yes" : "no"); } // Driver code to test abovea methods int main() { int edges[][2] = { { 0, 1 }, { 0, 2 }, { 1, 3 }, { 1, 4 }, { 2, 5 }, { 4, 6 }, { 5, 7 } }; int E = sizeof(edges) / sizeof(edges[0]); int V = E + 1; int timeIn[V], timeOut[V]; preProcess(edges, V, timeIn, timeOut); int u = 1; int v = 6; cout << isAncestor(u, v, timeIn, timeOut) << endl; u = 1; v = 7; cout << isAncestor(u, v, timeIn, timeOut) << endl; return 0; }

Output:

yes no

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