Query for ancestor-descendant relationship in a tree

3.5

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:

treeForAncestor1
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

This article is contributed by Utkarsh Trivedi. 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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