Euler Tour Technique

C++

const int MX = 2e5+5;
vector<int> adj[MX];
int timer = 0, st[MX], en[MX];

void dfs(int node, int parent) {
	st[node] = timer++;
	for (int i : adj[node]) {
		if (i != parent) {
			dfs(i, node);
		}
	}
	en[node] = timer-1;
}

Java

public static void dfs(int i, int p) {
	st[i] = timer++;
	for (int next : g[i]) {
		if(next != p) dfs(next, i);
	}
	en[i] = timer-1;
}

C++

/**
 * Description: 1D point update, range query where \texttt{comb} is
	* any associative operation. If $N=2^p$ then \texttt{seg[1]==query(0,N-1)}.
 * Time: O(\log N)
 * Source: 
	* http://codeforces.com/blog/entry/18051
	* KACTL
 * Verification: SPOJ Fenwick
 */

Java

import java.util.*;
import java.io.*;

public class Main {
	public static int[] st;
	public static int[] en;
	public static int timer, n;
	public static ArrayList<Integer> g[];
   
	//Segtree code

C++

int n; // The number of nodes in the graph
vector<int> graph[100000];
int timer = 0, tin[100000], euler_tour[200000];
int segtree[800000]; // Segment tree for RMQ

void dfs(int node = 0, int parent = -1) {
	tin[node] = timer; // The time when we first visit a node
	euler_tour[timer++] = node;
	for (int i : graph[node]) {
		if (i != parent) {

Java

import java.util.*;
import java.io.*;

public class LCA {
	public static int[] euler_tour, tin;
	public static int timer, size, N;
	public static ArrayList<Integer> g[];
   
	//Segtree code
	public static final int maxsize = (int)1e7;  // limit for array size

C++