Lowest Common Ancestor in a Binary Tree | Set 3 (Using RMQ)

Given a rooted tree, and two nodes are in the tree, find the Lowest common ancestor of both the nodes. The LCA for two nodes u and v is defined as the farthest node from the root that is the ancestor to both u and v. 
Prerequisites: LCA | SET 1

Example for the above figure :

Input : 4 5
Output : 2

Input : 4 7
Output : 1

Converting LCA to RMQ(Range Minimum Query):
Take an array named E[], which stores the order of dfs traversal i.e. the order in which the nodes are covered during the dfs traversal. For example, 

The tree given above has dfs traversal in the order: 1-2-4-2-5-2-1-3.
Take another array L[], in which L[i] is the level of node E[i].
And the array H[], which stores the index of the first occurrence of ith node in the array E[].
So, for the above tree, 
E[] = {1, 2, 4, 2, 5, 2, 1, 3} 
L[] = {1, 2, 3, 2, 3, 2, 1, 2} 
H[] = {0, 1, 7, 2, 4} 

Note that the arrays E and L are with one-based indexing but the array H has zero-based indexing.

Now, to find the LCA(4, 3), first, use the array H and find the indices at which 4 and 3 are found in E i.e. H[4] and H[3]. So, the indices come out to be 2 and 7. Now, look at the subarray L[2 : 7], and find the minimum in this subarray which is 1 (at the 6th index), and the corresponding element in the array E i.e. E[6] is the LCA(4, 3).

To understand why this works, take LCA(4, 3) again. The path by which one can reach node 3 from node 4 is the subarray E[2 : 7]. And, if there is a node with the lowest level in this path, then it can simply be claimed to be the LCA(4, 3).

Now, the problem is to find the minimum in the subarray E[H[u]….H[v]] (assuming that H[u] >= H[v]). And, that could be done using a segment tree or sparse table. Below is the code using the segment tree. 

Implementation:

C++

// CPP code to find LCA of given
// two nodes in a tree
#include <bits/stdc++.h>

#define sz(x) x.size()
#define pb push_back
#define left 2 * i + 1
#define right 2 * i + 2
using namespace std;

const int maxn = 100005;

// the graph
vector<vector<int>> g(maxn);

// level of each node
int level[maxn];

vector<int> e;
vector<int> l;
int h[maxn];

// the segment tree
int st[5 * maxn];

// adding edges to the graph(tree)
void add_edge(int u, int v) {
  g[u].pb(v);
  g[v].pb(u);
}

// assigning level to nodes
void leveling(int src) {
  for (int i = 0; i < sz(g[src]); i++) {
    int des = g[src][i];
    if (!level[des]) {
      level[des] = level[src] + 1;
      leveling(des);
    }
  }
}

bool visited[maxn];

// storing the dfs traversal
// in the array e
void dfs(int src) {
  e.pb(src);
  visited[src] = 1;
  for (int i = 0; i < sz(g[src]); i++) {
    int des = g[src][i];
    if (!visited[des]) {
      dfs(des);
      e.pb(src);
    }
  }
}

// making the array l
void setting_l(int n) {
  for (int i = 0; i < sz(e); i++)
    l.pb(level[e[i]]);
}

// making the array h
void setting_h(int n) {
  for (int i = 0; i <= n; i++)
    h[i] = -1;
  for (int i = 0; i < sz(e); i++) {
    // if is already stored
    if (h[e[i]] == -1)
      h[e[i]] = i;
  }
}

// Range minimum query to return the index
// of minimum in the subarray L[qs:qe]
int RMQ(int ss, int se, int qs, int qe, int i) {
  if (ss > se)
    return -1;

  // out of range
  if (se < qs || qe < ss)
    return -1;

  // in the range
  if (qs <= ss && se <= qe)
    return st[i];

  int mid = (ss + se) >> 1;
  int st = RMQ(ss, mid, qs, qe, left);
  int en = RMQ(mid + 1, se, qs, qe, right);

  if (st != -1 && en != -1) {
    if (l[st] < l[en])
      return st;
    return en;
  } else if (st != -1)
    return st;
  else if (en != -1)
    return en;
}

// constructs the segment tree
void SegmentTreeConstruction(int ss, int se, int i) {
  if (ss > se)
    return;
  if (ss == se) // leaf
  {
    st[i] = ss;
    return;
  }
  int mid = (ss + se) >> 1;

  SegmentTreeConstruction(ss, mid, left);
  SegmentTreeConstruction(mid + 1, se, right);

  if (l[st[left]] < l[st[right]])
    st[i] = st[left];
  else
    st[i] = st[right];
}

// Function to get LCA
int LCA(int x, int y) {
  if (h[x] > h[y])
    swap(x, y);
  return e[RMQ(0, sz(l) - 1, h[x], h[y], 0)];
}

// Driver code
int main() {

  // n=number of nodes in the tree
  // q=number of queries to answer
  int n = 15, q = 5;

  // making the tree
  /*
                   1
                 / | \
                2  3  4
                   |   \
                   5    6
                 / |  \
               8   7    9 (right of 5)
                 / | \   | \
               10 11 12 13 14
                      |
                      15
  */
  add_edge(1, 2);
  add_edge(1, 3);
  add_edge(1, 4);
  add_edge(3, 5);
  add_edge(4, 6);
  add_edge(5, 7);
  add_edge(5, 8);
  add_edge(5, 9);
  add_edge(7, 10);
  add_edge(7, 11);
  add_edge(7, 12);
  add_edge(9, 13);
  add_edge(9, 14);
  add_edge(12, 15);

  level[1] = 1;
  leveling(1);

  dfs(1);

  setting_l(n);

  setting_h(n);

  SegmentTreeConstruction(0, sz(l) - 1, 0);

  cout << LCA(10, 15) << endl;
  cout << LCA(11, 14) << endl;

  return 0;
}

Java

// JAVA code to find LCA of given
// two nodes in a tree
import java.util.*;
public class GFG
{

  static int maxn = 100005;
  static  int left(int i)
  {
    return  (2 * i + 1);
  }
  static  int right(int i) { return 2 * i + 2;}

  // the graph
  static Vector<Integer> []g = new Vector[maxn];

  // level of each node
  static int []level = new int[maxn];
  static Vector<Integer> e = new Vector<>();
  static Vector<Integer> l= new Vector<>();
  static int []h = new int[maxn];

  // the segment tree
  static int []st = new int[5 * maxn];

  // adding edges to the graph(tree)
  static void add_edge(int u, int v)
  {
    g[u].add(v);
    g[v].add(u);
  }

  // assigning level to nodes
  static void levelling(int src)
  {
    for (int i = 0; i < (g[src].size()); i++)
    {
      int des = g[src].get(i);
      if (level[des] != 0)
      {
        level[des] = level[src] + 1;
        leveling(des);
      }
    }
  }

  static boolean []visited = new boolean[maxn];

  // storing the dfs traversal
  // in the array e
  static void dfs(int src)
  {
    e.add(src);
    visited[src] = true;
    for (int i = 0; i < (g[src]).size(); i++)
    {
      int des = g[src].get(i);
      if (!visited[des])
      {
        dfs(des);
        e.add(src);
      }
    }
  }

  // making the array l
  static void setting_l(int n)
  {
    for (int i = 0; i < e.size(); i++)
      l.add(level[e.get(i)]);
  }

  // making the array h
  static void setting_h(int n)
  {
    for (int i = 0; i <= n; i++)
      h[i] = -1;
    for (int i = 0; i < e.size(); i++)
    {

      // if is already stored
      if (h[e.get(i)] == -1)
        h[e.get(i)] = i;
    }
  }

  // Range minimum query to return the index
  // of minimum in the subarray L[qs:qe]
  static int RMQ(int ss, int se, int qs, int qe, int i)
  {
    if (ss > se)
      return -1;

    // out of range
    if (se < qs || qe < ss)
      return -1;

    // in the range
    if (qs <= ss && se <= qe)
      return st[i];

    int mid = (ss + se)/2 ;
    int st = RMQ(ss, mid, qs, qe, left(i));
    int en = RMQ(mid + 1, se, qs, qe, right(i));

    if (st != -1 && en != -1)
    {
      if (l.get(st) < l.get(en))
        return st;
      return en;
    } else if (st != -1)
      return st-2;
    else if (en != -1)
      return en-1;
    return 0;
  }

  // constructs the segment tree
  static void SegmentTreeConstruction(int ss,
                                      int se, int i)
  {
    if (ss > se)
      return;
    if (ss == se) // leaf
    {
      st[i] = ss;
      return;
    }
    int mid = (ss + se) /2;

    SegmentTreeConstruction(ss, mid, left(i));
    SegmentTreeConstruction(mid + 1, se, right(i));
    if (l.get(st[left(i)]) < l.get(st[right(i)]))
      st[i] = st[left(i)];
    else
      st[i] = st[right(i)];
  }

  // Function to get LCA
  static int LCA(int x, int y)
  {
    if (h[x] > h[y])
    {
      int t = x;
      x = y;
      y = t;
    }
    return e.get(RMQ(0, l.size() - 1, h[x], h[y], 0));
  }

  // Driver code
  public static void main(String[] args)
  {

    // n=number of nodes in the tree
    // q=number of queries to answer
    int n = 15, q = 5;
    for (int i = 0; i < g.length; i++)
      g[i] = new Vector<Integer>();

    // making the tree
    /*
                   1
                 / | \
                2  3  4
                   |   \
                   5    6
                 / |  \
               8   7    9 (right of 5)
                 / | \   | \
               10 11 12 13 14
                      |
                      15
  */
    add_edge(1, 2);
    add_edge(1, 3);
    add_edge(1, 4);
    add_edge(3, 5);
    add_edge(4, 6);
    add_edge(5, 7);
    add_edge(5, 8);
    add_edge(5, 9);
    add_edge(7, 10);
    add_edge(7, 11);
    add_edge(7, 12);
    add_edge(9, 13);
    add_edge(9, 14);
    add_edge(12, 15);
    level[1] = 1;
    leveling(1);
    dfs(1);
    setting_l(n);
    setting_h(n);
    SegmentTreeConstruction(0, l.size() - 1, 0);
    System.out.print(LCA(10, 15) +"\n");
    System.out.print(LCA(11, 14) +"\n");
  }
}

// This code is contributed by Rajput-Ji

Python3

# Python code to find LCA of given
# two nodes in a tree

maxn = 100005

# the graph
g = [[] for i in range(maxn)]

# level of each node
level = [0] * maxn

e = []
l = []
h = [0] * maxn

# the segment tree
st = [0] * (5 * maxn)

# adding edges to the graph(tree)
def add_edge(u: int, v: int):
    g[u].append(v)
    g[v].append(u)

# assigning level to nodes
def levelling(src: int):
    for i in range(len(g[src])):
        des = g[src][i]
        if not level[des]:
            level[des] = level[src] + 1
            leveling(des)

visited = [False] * maxn

# storing the dfs traversal
# in the array e
def dfs(src: int):
    e.append(src)
    visited[src] = True
    for i in range(len(g[src])):
        des = g[src][i]
        if not visited[des]:
            dfs(des)
            e.append(src)

# making the array l
def setting_l(n: int):
    for i in range(len(e)):
        l.append(level[e[i]])

# making the array h
def setting_h(n: int):
    for i in range(n + 1):
        h[i] = -1
    for i in range(len(e)):

        # if is already stored
        if h[e[i]] == -1:
            h[e[i]] = i

# Range minimum query to return the index
# of minimum in the subarray L[qs:qe]
def RMQ(ss: int, se: int, qs: int, qe: int, i: int) -> int:
    global st
    if ss > se:
        return -1

    # out of range
    if se < qs or qe < ss:
        return -1

    # in the range
    if qs <= ss and se <= qe:
        return st[i]

    mid = (se + ss) >> 1
    stt = RMQ(ss, mid, qs, qe, 2 * i + 1)
    en = RMQ(mid + 1, se, qs, qe, 2 * i + 2)

    if stt != -1 and en != -1:
        if l[stt] < l[en]:
            return stt
        return en
    elif stt != -1:
        return stt
    elif en != -1:
        return en

# constructs the segment tree
def segmentTreeConstruction(ss: int, se: int, i: int):
    if ss > se:
        return
    if ss == se: # leaf
        st[i] = ss
        return

    mid = (ss + se) >> 1
    segmentTreeConstruction(ss, mid, 2 * i + 1)
    segmentTreeConstruction(mid + 1, se, 2 * i + 2)

    if l[st[2 * i + 1]] < l[st[2 * i + 2]]:
        st[i] = st[2 * i + 1]
    else:
        st[i] = st[2 * i + 2]

# Function to get LCA
def LCA(x: int, y: int) -> int:
    if h[x] > h[y]:
        x, y = y, x
    return e[RMQ(0, len(l) - 1, h[x], h[y], 0)]

# Driver Code
if __name__ == "__main__":

    # n=number of nodes in the tree
    # q=number of queries to answer
    n = 15
    q = 5

    # making the tree
    # /*
    #         1
    #     / | \
    #     2 3 4
    #         | \
    #         5 6
    #     / | \
    #     8 7 9 (right of 5)
    #     / | \ | \
    #     10 11 12 13 14
    #             |
    #             15
    # */
    add_edge(1, 2)
    add_edge(1, 3)
    add_edge(1, 4)
    add_edge(3, 5)
    add_edge(4, 6)
    add_edge(5, 7)
    add_edge(5, 8)
    add_edge(5, 9)
    add_edge(7, 10)
    add_edge(7, 11)
    add_edge(7, 12)
    add_edge(9, 13)
    add_edge(9, 14)
    add_edge(12, 15)

    level[1] = 1
    leveling(1)
    dfs(1)
    setting_l(n)
    setting_h(n)

    segmentTreeConstruction(0, len(l) - 1, 0)

    print(LCA(10, 15))
    print(LCA(11, 14))

# This code is contributed by
# sanjeev2552

C#

// C# code to find LCA of given
// two nodes in a tree
using System;
using System.Collections.Generic;
public class GFG
{
  static int maxn = 100005;
  static  int left(int i)
  {
    return  (2 * i + 1);
  }
  static  int right(int i) { return 2 * i + 2;}

  // the graph
  static List<int> []g = new List<int>[maxn];

  // level of each node
  static int []level = new int[maxn];
  static List<int> e = new List<int>();
  static List<int> l= new List<int>();
  static int []h = new int[maxn];

  // the segment tree
  static int []st;

  // adding edges to the graph(tree)
  static void add_edge(int u, int v)
  {
    g[u].Add(v);
    g[v].Add(u);
  }

  // assigning level to nodes
  static void leveling(int src)
  {
    for (int i = 0; i < (g[src].Count); i++)
    {
      int des = g[src][i];
      if (level[des] != 0)
      {
        level[des] = level[src] + 1;
        leveling(des);
      }
    }
  }
  static bool []visited = new bool[maxn];

  // storing the dfs traversal
  // in the array e
  static void dfs(int src)
  {
    e.Add(src);
    visited[src] = true;
    for (int i = 0; i < (g[src]).Count; i++)
    {
      int des = g[src][i];
      if (!visited[des])
      {
        dfs(des);
        e.Add(src);
      }
    }
  }

  // making the array l
  static void setting_l(int n)
  {
    for (int i = 0; i < e.Count; i++)
      l.Add(level[e[i]]);
  }

  // making the array h
  static void setting_h(int n)
  {
    for (int i = 0; i <= n; i++)
      h[i] = -1;
    for (int i = 0; i < e.Count; i++)
    {

      // if is already stored
      if (h[e[i]] == -1)
        h[e[i]] = i;
    }
  }

  // Range minimum query to return the index
  // of minimum in the subarray L[qs:qe]
  static int RMQ(int ss, int se, int qs, int qe, int i)
  {
    if (ss > se)
      return -1;

    // out of range
    if (se < qs || qe < ss)
      return -1;

    // in the range
    if (qs <= ss && se <= qe)
      return st[i];

    int mid = (ss + se)/2 ;
    int sti = RMQ(ss, mid, qs, qe, left(i));
    int en = RMQ(mid + 1, se, qs, qe, right(i));

    if (sti != -1 && en != -1)
    {
      if (l[sti] < l[en])
        return sti;
      return en;
    } else if (sti != -1)
      return sti-2;
    else if (en != -1)
      return en-1;
    return 0;
  }

  // constructs the segment tree
  static void SegmentTreeConstruction(int ss,
                                      int se, int i)
  {
    if (ss > se)
      return;
    if (ss == se) // leaf
    {
      st[i] = ss;
      return;
    }
    int mid = (ss + se) /2;

    SegmentTreeConstruction(ss, mid, left(i));
    SegmentTreeConstruction(mid + 1, se, right(i));
    if (l[st[left(i)]] < l[st[right(i)]])
      st[i] = st[left(i)];
    else
      st[i] = st[right(i)];
  }

  // Function to get LCA
  static int LCA(int x, int y)
  {
    if (h[x] > h[y])
    {
      int t = x;
      x = y;
      y = t;
    }
    return e[RMQ(0, l.Count - 1, h[x], h[y], 0)];
  }

  // Driver code
  public static void Main(String[] args)
  {
    st = new int[5 * maxn];

    // n=number of nodes in the tree
    // q=number of queries to answer
    int n = 15;
    for (int i = 0; i < g.Length; i++)
      g[i] = new List<int>();

    // making the tree
    /*
                   1
                 / | \
                2  3  4
                   |   \
                   5    6
                 / |  \
               8   7    9 (right of 5)
                 / | \   | \
               10 11 12 13 14
                      |
                      15
  */
    add_edge(1, 2);
    add_edge(1, 3);
    add_edge(1, 4);
    add_edge(3, 5);
    add_edge(4, 6);
    add_edge(5, 7);
    add_edge(5, 8);
    add_edge(5, 9);
    add_edge(7, 10);
    add_edge(7, 11);
    add_edge(7, 12);
    add_edge(9, 13);
    add_edge(9, 14);
    add_edge(12, 15);
    level[1] = 1;
    leveling(1);
    dfs(1);
    setting_l(n);
    setting_h(n);
    SegmentTreeConstruction(0, l.Count - 1, 0);
    Console.Write(LCA(10, 15) +"\n");
    Console.Write(LCA(11, 14) +"\n");
  }
}

// This code is contributed by gauravrajput1

Javascript

<script>

    // JavaScript code to find LCA of given
    // two nodes in a tree

    let maxn = 100005;
    function left(i)
    {
      return  (2 * i + 1);
    }
    function right(i) { return 2 * i + 2;}

    // the graph
    let g = new Array(maxn);

    // level of each node
    let level = new Array(maxn);
    level.fill(0);
    let e = [];
    let l= [];
    let h = new Array(maxn);
    h.fill(0);

    // the segment tree
    let st = new Array(5 * maxn);
    st.fill(0);

    // adding edges to the graph(tree)
    function add_edge(u, v)
    {
      g[u].push(v);
      g[v].push(u);
    }

    // assigning level to nodes
    function levelling(src)
    {
      for (let i = 0; i < (g[src].length); i++)
      {
        let des = g[src][i];
        if (level[des] != 0)
        {
          level[des] = level[src] + 1;
          levelling(des);
        }
      }
    }

    let visited = new Array(maxn);
    visited.fill(false);

    // storing the dfs traversal
    // in the array e
    function dfs(src)
    {
      e.push(src);
      visited[src] = true;
      for (let i = 0; i < (g[src]).length; i++)
      {
        let des = g[src][i];
        if (!visited[des])
        {
          dfs(des);
          e.push(src);
        }
      }
    }

    // making the array l
    function setting_l(n)
    {
      for (let i = 0; i < e.length; i++)
        l.push(level[e[i]]);
    }

    // making the array h
    function setting_h(n)
    {
      for (let i = 0; i <= n; i++)
        h[i] = -1;
      for (let i = 0; i < e.length; i++)
      {

        // if is already stored
        if (h[e[i]] == -1)
          h[e[i]] = i;
      }
    }

    // Range minimum query to return the index
    // of minimum in the subarray L[qs:qe]
    function RMQ(ss, se, qs, qe, i)
    {
      if (ss > se)
        return -1;

      // out of range
      if (se < qs || qe < ss)
        return -1;

      // in the range
      if (qs <= ss && se <= qe)
        return st[i];

      let mid = parseInt((ss + se)/2 , 10);
      let St = RMQ(ss, mid, qs, qe, left(i));
      let en = RMQ(mid + 1, se, qs, qe, right(i));

      if (St != -1 && en != -1)
      {
        if (l[St] < l[en])
          return St;
        return en;
      } else if (St != -1)
        return St-2;
      else if (en != -1)
        return en-1;
      return 0;
    }

    // constructs the segment tree
    function SegmentTreeConstruction(ss, se, i)
    {
      if (ss > se)
        return;
      if (ss == se) // leaf
      {
        st[i] = ss;
        return;
      }
      let mid = parseInt((ss + se) /2, 10);

      SegmentTreeConstruction(ss, mid, left(i));
      SegmentTreeConstruction(mid + 1, se, right(i));
      if (l[st[left(i)]] < l[st[right(i)]])
        st[i] = st[left(i)];
      else
        st[i] = st[right(i)];
    }

    // Function to get LCA
    function LCA(x, y)
    {
      if (h[x] > h[y])
      {
        let t = x;
        x = y;
        y = t;
      }
      return e[RMQ(0, l.length - 1, h[x], h[y], 0)];
    }

    // n=number of nodes in the tree
    // q=number of queries to answer
    let n = 15, q = 5;
    for (let i = 0; i < g.length; i++)
      g[i] = [];

    // making the tree
    /*
                   1
                 / | \
                2  3  4
                   |   \
                   5    6
                 / |  \
               8   7    9 (right of 5)
                 / | \   | \
               10 11 12 13 14
                      |
                      15
  */
    add_edge(1, 2);
    add_edge(1, 3);
    add_edge(1, 4);
    add_edge(3, 5);
    add_edge(4, 6);
    add_edge(5, 7);
    add_edge(5, 8);
    add_edge(5, 9);
    add_edge(7, 10);
    add_edge(7, 11);
    add_edge(7, 12);
    add_edge(9, 13);
    add_edge(9, 14);
    add_edge(12, 15);
    level[1] = 1;
    levelling(1);
    dfs(1);
    setting_l(n);
    setting_h(n);
    SegmentTreeConstruction(0, l.length - 1, 0);
    document.write(LCA(10, 15) +"</br>");
    document.write(LCA(11, 14) +"</br>");

</script>

Output

7
5

Time Complexity: The arrays defined are stored in O(n). The segment tree construction also takes O(n) time. The LCA function calls the function RMQ which takes O(logn) per query (as it uses the segment tree). So overall time complexity is O(n + q * logn).
Auxiliary Space: O(n)