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

Given a rooted tree, and two nodes which 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

lca

Example for 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.

C++

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// CPP code to find LCA of given
// two nodes in a tree
#include <algorithm>
#include <iostream>
#include <vector>
  
#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];
}
  
// Funtion 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() {
  ios::sync_with_stdio(0);
  
  // 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;
}

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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).



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