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Query for ancestor-descendant relationship in a tree

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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;
}


Java




// Java program to query whether two node has
// ancestor-descendant relationship or not
import java.util.Vector;
 
class GFG
{
  static int cnt;
 
  // Utility dfs method to assign in and out time
  // to each node
  static void dfs(Vector<Integer> []g, int u, int parent,
                  int []timeIn, int []timeOut)
  {
 
    // 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].get(i);       
      if (v != parent)
        dfs(g, v, u, timeIn, timeOut);
    }
 
    // Assign Out-time to node u
    timeOut[u] = cnt++;
  }
 
  // Method to preprocess all nodes for assigning time
  static void preProcess(int [][]edges, int V,
                         int []timeIn, int []timeOut)
  {
    @SuppressWarnings("unchecked")
    Vector<Integer> []g = new Vector[V];
    for (int i = 0; i < g.length; i++)
      g[i] = new Vector<Integer>();
 
    // Conarray 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].add(v);
      g[v].add(u);
    }
    cnt = 0;
 
    // Call dfs method from root
    dfs(g, 0, -1, timeIn, timeOut);
  }
 
  // Method returns "yes" if u is a ancestor
  // node of v
  static String isAncestor(int u, int v, int []timeIn,
                           int []timeOut)
  {
    boolean b = (timeIn[u] <= timeIn[v] &&
                 timeOut[v] <= timeOut[u]);
    return (b ? "yes" : "no");
  }
 
  // Driver code   
  public static void main(String[] args)
  {
    int edges[][] = { { 0, 1 },
                     { 0, 2 },
                     { 1, 3 },
                     { 1, 4 },
                     { 2, 5 },
                     { 4, 6 },
                     { 5, 7 } };   
    int E = edges.length;
    int V = E + 1;  
    int []timeIn = new int[V];
    int []timeOut = new int[V];  
    preProcess(edges, V, timeIn, timeOut);   
    int u = 1;
    int v = 6;
    System.out.println(isAncestor(u, v, timeIn,
                                  timeOut));   
    u = 1;
    v = 7;
    System.out.println(isAncestor(u, v, timeIn,
                                  timeOut));
  }
}
 
// This code is contributed by gauravrajput1


Python3




# Python program to query whether two node has
# ancestor-descendant relationship or not
 
cnt = 0
 
# Utility dfs method to assign in and out time
# to each node
def dfs(g: list, u: int, parent: int, timeIn: list, timeOut: list):
    global cnt
 
    # assign In-time to node u
    timeIn[u] = cnt
    cnt += 1
 
    # call dfs over all neighbors except parent
    for i in range(len(g[u])):
        v = g[u][i]
        if v != parent:
            dfs(g, v, u, timeIn, timeOut)
 
    # assign Out-time to node u
    timeOut[u] = cnt
    cnt += 1
 
# method to preprocess all nodes for assigning time
def preProcess(edges: list, V: int, timeIn: list, timeOut: list):
    global cnt
    g = [[] for i in range(V)]
 
    # construct array of vector data structure
    # for tree
    for i in range(V - 1):
        u = edges[i][0]
        v = edges[i][1]
 
        g[u].append(v)
        g[v].append(u)
 
    cnt = 0
 
    # call dfs method from root
    dfs(g, 0, -1, timeIn, timeOut)
 
# method returns "yes" if u is a ancestor
# node of v
def isAncestor(u: int, v: int, timeIn: list, timeOut: list) -> str:
    b = timeIn[u] <= timeIn[u] and timeOut[v] <= timeOut[u]
    return "yes" if b else "no"
 
# Driver Code
if __name__ == "__main__":
    edges = [(0, 1), (0, 2), (1, 3), (1, 4), (2, 5), (4, 6), (5, 7)]
    E = len(edges)
    V = E + 1
 
    timeIn = [0] * V
    timeOut = [0] * V
    preProcess(edges, V, timeIn, timeOut)
 
    u = 1
    v = 6
    print(isAncestor(u, v, timeIn, timeOut))
 
    u = 1
    v = 7
    print(isAncestor(u, v, timeIn, timeOut))
 
# This code is contributed by
# sanjeev2552


C#




// C# program to query whether two node has
// ancestor-descendant relationship or not
using System;
using System.Collections;
 
class GFG{
 
// Utility dfs method to assign in and out time
// to each node
static void dfs(ArrayList []g, int u, int parent,
                int []timeIn, int []timeOut,
                ref 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].Count; i++)
    {
        int v = (int)g[u][i];
         
        if (v != parent)
            dfs(g, v, u, timeIn, timeOut, ref cnt);
    }
  
    // Assign Out-time to node u
    timeOut[u] = cnt++;
}
  
// Method to preprocess all nodes for assigning time
static void preProcess(int [,]edges, int V,
                       int []timeIn, int []timeOut)
{
    ArrayList []g = new ArrayList[V];
     
    for(int i = 0; i < V; i++)
    {
        g[i] = new ArrayList();
    }
  
    // 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].Add(v);
        g[v].Add(u);
    }
  
    int cnt = 0;
  
    // Call dfs method from root
    dfs(g, 0, -1, timeIn, timeOut, ref cnt);
}
  
// Method returns "yes" if u is a ancestor
// node of v
static 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   
static void Main()
{
    int [,]edges = { { 0, 1 },
                     { 0, 2 },
                     { 1, 3 },
                     { 1, 4 },
                     { 2, 5 },
                     { 4, 6 },
                     { 5, 7 } };
     
    int E = edges.GetLength(0);
    int V = E + 1;
     
    int []timeIn = new int[V];
    int []timeOut = new int[V];
     
    preProcess(edges, V, timeIn, timeOut);
     
    int u = 1;
    int v = 6;
    Console.Write(isAncestor(u, v, timeIn,
                             timeOut) + "\n");
     
    u = 1;
    v = 7;
    Console.Write(isAncestor(u, v, timeIn,
                             timeOut) + "\n");
}
}
 
// This code is contributed by rutvik_56


Javascript




<script>
 
// Javascript program to query whether
// two node has ancestor-descendant
// relationship or not
let cnt;
 
// Utility dfs method to assign in and out time
// to each node
function dfs(g, u, parent, timeIn, timeOut)
{
     
    // Assign In-time to node u
    timeIn[u] = cnt++;
     
    // Call dfs over all neighbors except parent
    for(let i = 0; i < g[u].length; i++)
    {
        let v = g[u][i];      
        if (v != parent)
            dfs(g, v, u, timeIn, timeOut);
    }
     
    // Assign Out-time to node u
    timeOut[u] = cnt++;
}
 
// Method to preprocess all nodes for assigning time
function preProcess(edges, V, timeIn, timeOut)
{
    let g = new Array(V);
    for(let i = 0; i < g.length; i++)
        g[i] = [];
     
    // Conarray of vector data structure
    // for tree
    for(let i = 0; i < V - 1; i++)
    {
        let u = edges[i][0];
        let v = edges[i][1];
        g[u].push(v);
        g[v].push(u);
    }
    cnt = 0;
     
    // Call dfs method from root
    dfs(g, 0, -1, timeIn, timeOut);
}
 
// Method returns "yes" if u is a ancestor
// node of v
function isAncestor(u, v, timeIn, timeOut)
{
    let b = (timeIn[u] <= timeIn[v] &&
             timeOut[v] <= timeOut[u]);
    return (b ? "yes" : "no");
}
 
// Driver code
let edges = [ [ 0, 1 ], [ 0, 2 ],
              [ 1, 3 ], [ 1, 4 ],
              [ 2, 5 ], [ 4, 6 ],
              [ 5, 7 ] ];  
let E = edges.length;
let V = E + 1; 
let timeIn = new Array(V);
let timeOut = new Array(V); 
preProcess(edges, V, timeIn, timeOut);
 
let u = 1;
let v = 6;
document.write(isAncestor(u, v, timeIn,
                          timeOut) + "</br>");  
u = 1;
v = 7;
document.write(isAncestor(u, v, timeIn,
                          timeOut));
                           
// This code is contributed by divyeshrabadiya07
 
</script>


Output: 

yes
no

Time complexity : O(N)

Auxiliary Space : O(N)

 



Last Updated : 10 Mar, 2023
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