Lowest Common Ancestor for a Set of Nodes in a Rooted Tree

Given a rooted tree with N nodes, the task is to find the Lowest Common Ancestor for a given set of nodes V of that tree.

Examples:

Input:    
                   1
                /  |  \
               2   3   4
             /  \  |   |
            5   6  7   10
                  / \
                 8   9
V[] = {7, 3, 8, 9}
Output: 3

Input:   
                   1
                /  |  \
               2   3   4
             /  \  |   |
            5   6  7   10
                  / \
                 8   9
V[] = {4, 6, 7}
Output: 1

Approach: We can observe that

  1. the Lowest Common Ancestors for any set of nodes will have in-time less than or equal to that of all nodes in the set and out-time greater than or equal(equal if LCA is present in the set) to that of all nodes in the set.
  2. Thus, in order to solve the problem, we need to traverse the entire tree starting from the root node using Depth First Search Traversal and store the in-time, out-time, and level for every node.
  3. The node with in-time less than or equal to all the nodes in the set and out-time greater than or equal to all the nodes in the set and with maximum level possible is the answer.

Below is the implementation of the above approach:

C++

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// C++ Program to find the LCA
// in a rooted tree for a given
// set of nodes
  
#include <bits/stdc++.h>
using namespace std;
  
// Set time 1 initially
int T = 1;
void dfs(int node, int parent,
         vector<int> g[],
         int level[], int t_in[],
         int t_out[])
{
  
    // Case for root node
    if (parent == -1) {
        level[node] = 1;
    }
    else {
        level[node] = level[parent] + 1;
    }
  
    // In-time for node
    t_in[node] = T;
    for (auto i : g[node]) {
        if (i != parent) {
            T++;
            dfs(i, node, g,
                level, t_in, t_out);
        }
    }
    T++;
  
    // Out-time for the node
    t_out[node] = T;
}
  
int findLCA(int n, vector<int> g[],
            vector<int> v)
{
  
    // level[i]--> Level of node i
    int level[n + 1];
  
    // t_in[i]--> In-time of node i
    int t_in[n + 1];
  
    // t_out[i]--> Out-time of node i
    int t_out[n + 1];
  
    // Fill the level, in-time
    // and out-time of all nodes
    dfs(1, -1, g,
        level, t_in, t_out);
  
    int mint = INT_MAX, maxt = INT_MIN;
    int minv = -1, maxv = -1;
    for (auto i = v.begin(); i != v.end(); i++) {
        // To find minimum in-time among
        // all nodes in LCA set
        if (t_in[*i] < mint) {
            mint = t_in[*i];
            minv = *i;
        }
        // To find maximum in-time among
        // all nodes in LCA set
        if (t_out[*i] > maxt) {
            maxt = t_out[*i];
            maxv = *i;
        }
    }
  
    // Node with same minimum
    // and maximum out time
    // is LCA for the set
    if (minv == maxv) {
        return minv;
    }
  
    // Take the minimum level as level of LCA
    int lev = min(level[minv], level[maxv]);
    int node, l = INT_MIN;
    for (int i = 1; i <= n; i++) {
        // If i-th node is at a higher level
        // than that of the minimum among
        // the nodes of the given set
        if (level[i] > lev)
            continue;
  
        // Compare in-time, out-time
        // and level of i-th node
        // to the respective extremes
        // among all nodes of the given set
        if (t_in[i] <= mint
            && t_out[i] >= maxt
            && level[i] > l) {
            node = i;
            l = level[i];
        }
    }
  
    return node;
}
  
// Driver code
int main()
{
    int n = 10;
    vector<int> g[n + 1];
    g[1].push_back(2);
    g[2].push_back(1);
    g[1].push_back(3);
    g[3].push_back(1);
    g[1].push_back(4);
    g[4].push_back(1);
    g[2].push_back(5);
    g[5].push_back(2);
    g[2].push_back(6);
    g[6].push_back(2);
    g[3].push_back(7);
    g[7].push_back(3);
    g[4].push_back(10);
    g[10].push_back(4);
    g[8].push_back(7);
    g[7].push_back(8);
    g[9].push_back(7);
    g[7].push_back(9);
  
    vector<int> v = { 7, 3, 8 };
  
    cout << findLCA(n, g, v) << endl;
}

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Output:

3

Time Complexity: O(N)
Space Complexity: O(N)

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