Find the Kth node in the DFS traversal of a given subtree in a Tree

Given a tree with N nodes, and two integers K and V. The task is to find the Kth node in the DFS traversal of the vertex V.

Consider the below Tree:

DFS of node number 1 is [1, 2, 3, 5, 6, 8, 7, 9, 4].
DFS of node number 3 is [3, 5, 6, 8, 7, 9]
DFS of node number 7 is [7, 9]
DFS of node number 9 is [9].

Print “-1” if the numbers in the DFS of vertex V are less than K.

Examples:

Input : Tree: Shown in above image, V = 3, K = 4
Output : 8

Input : Tree: Shown in above image, V = 7, K = 3
Output : -1

Approach : Let’s construct a vector p: to store the DFS traversal of the complete tree from vertex 1. Let tinv be the position of the vertex V in the vector p (the size of the vector p in moment we call DFS from the vertex V) and toutv be the position of the first vertex pushed to the vector after leaving the subtree of vertex V (the size of the vector p in moment when we return from DFS from the vertex V). Then it is obvious that the subtree of the vertex V lies in the interval [tinv, toutv).

So, to find the Kth node in the DFS of the subtree of node V, we will have to return the Kth node in the interval [tinv, toutv).

Below is the implementation of the above approach:

C++

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// C++ program to find the Kth node in the
// DFS traversal of the subtree of given
// vertex V in a Tree
  
#include <bits/stdc++.h>
using namespace std;
#define N 100005
  
// To store nodes
int n;
vector<int> tree[N];
  
// To store the current index of vertex in DFS
int currentIdx;
  
// To store the starting index and ending
// index of vertex in the DFS traversal array
vector<int> startIdx, endIdx;
  
// To store the DFS of vertex 1
vector<int> p;
  
// Function to add edge between two nodes
void Add_edge(int u, int v)
{
    tree[u].push_back(v);
    tree[v].push_back(u);
}
  
// Initialize the vectors
void intisalise()
{
    startIdx.resize(n);
    endIdx.resize(n);
    p.resize(n);
}
  
// Function to perform DFS of a vertex
// 1. stores the DFS of the vertex 1 in vector p,
// 2. store the start index of DFS of every vertex
// 3. store the end index of DFS of every vertex
void Dfs(int ch, int par)
{
    p[currentIdx] = ch;
  
    // store staring index of node ch
    startIdx[ch] = currentIdx++;
  
    for (auto c : tree[ch]) {
        if (c != par)
            Dfs(c, ch);
    }
  
    // store ending index
    endIdx[ch] = currentIdx - 1;
}
  
// Function to find the Kth node in DFS of vertex V
int findNode(int v, int k)
{
    k += startIdx[v] - 1;
  
    // check if kth number exits or not
    if (k <= endIdx[v])
        return p[k];
  
    return -1;
}
  
// Driver code
int main()
{
    // number of nodes
    n = 9;
  
    // add edges
    Add_edge(1, 2);
    Add_edge(1, 3);
    Add_edge(1, 4);
    Add_edge(3, 5);
    Add_edge(3, 7);
    Add_edge(5, 6);
    Add_edge(5, 8);
    Add_edge(7, 9);
  
    intisalise();
  
    // store DFS of 1st node
    Dfs(1, 0);
  
    int v = 3, k = 4;
  
    cout << findNode(v, k);
  
    return 0;
}

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Python3

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# Python3 program to find the Kth node in the
# DFS traversal of the subtree of given
# vertex V in a Tree
N = 100005
  
n = 10
tree = [[]for i in range(N)]
  
# To store the current index of vertex in DFS
currentIdx = 0
  
# To store the starting index and ending
# index of vertex in the DFS traversal array
startIdx = [0 for i in range(n)]
endIdx = [0 for i in range(n)]
  
# To store the DFS of vertex 1
p = [0 for i in range(n)]
  
# Function to add edge between two nodes
def Add_edge(u, v):
    tree[u].append(v)
    tree[v].append(u)
  
# Initialize the vectors
def intisalise():
    pass
  
# Function to perform DFS of a vertex
# 1. stores the DFS of the vertex 1 in vector p,
# 2. store the start index of DFS of every vertex
# 3. store the end index of DFS of every vertex
def Dfs(ch, par):
    global currentIdx
  
    p[currentIdx] = ch
  
    # store staring index of node ch
    startIdx[ch] = currentIdx
    currentIdx += 1
  
    for c in tree[ch]:
        if (c != par):
            Dfs(c, ch)
  
    # store ending index
    endIdx[ch] = currentIdx - 1
  
# Function to find the Kth node in 
# DFS of vertex V
def findNode(v, k):
  
    k += startIdx[v] - 1
  
    # check if kth number exits or not
    if (k <= endIdx[v]):
        return p[k]
  
    return -1
  
# Driver code
  
# number of nodes
n = 9
  
# add edges
Add_edge(1, 2)
Add_edge(1, 3)
Add_edge(1, 4)
Add_edge(3, 5)
Add_edge(3, 7)
Add_edge(5, 6)
Add_edge(5, 8)
Add_edge(7, 9)
  
intisalise()
  
# store DFS of 1st node
Dfs(1, 0)
  
v, k = 3, 4
  
print(findNode(v, k))
  
# This code is contributed by mohit kumar

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Output:

8


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