Find LCA for K queries in Complete Binary Tree

By Ann Roberts On Jan 19, 2023

Given an integer n. There is a complete binary tree with 2ⁿ – 1 nodes. The root of that tree is the node with the value 1, and every node with a value x has two children where the left node has the value
2*x and the right node has the value 2*x + 1, you are given K queries of type (a_i, b_i), and the task is to return the LCA for the node pair a_iand b_ifor all K queries.

Examples:

Input: n = 5, queries = [ { 17, 21 }, { 23, 5 }, { 15, 7 }, { 3, 21 }, { 31, 9 }, { 5, 15 }, { 11, 2 }, { 19, 7 } ]

Complete binary tree for given input n=5

Output: [ 2, 5, 7, 1, 1, 1, 2, 1 ]

Input: n = 3, queries = [ {2, 5}, {3, 6}, {4, 1}, {7, 3} ]

Complete binary tree for given input n=3

Output: [2, 3, 1, 3]

Approach: The problem can be solved based on the following idea:

As all values on a level are smaller than values on the next level. Check which node is having greater value in a query, and divide it by 2 to reach its parent node. Repeat this step until we get common element.

Follow the steps to solve the problem:

In a query, we are having 2 nodes a and b, whose lowest common ancestor we have to find.
By dividing the value of the node by 2, we will always get the parent node value.
From a and b whichever node is having greater value divide by 2. So, as to move towards the root of the root.
When a and b becomes equal, the common ancestor between them is got and returned.

Below is the implementation for the approach discussed:

C++

#include <bits/stdc++.h>

using namespace std;

int helper(int a, int b)

{

while (a != b) {

if (a > b)

a = a / 2;

else

b = b / 2;

}

return a;

}

int main()

{

int n = 5;

vector<vector<int> > queries

= { { 17, 21 }, { 23, 5 }, { 15, 7 }, { 3, 21 }, { 31, 9 }, { 5, 15 }, { 11, 2 }, { 19, 7 } };

for (auto e : queries) {

int lca = helper(e[0], e[1]);

cout << lca << ' ';

}

return 0;

}

Time Complexity: O(n)
Auxiliary Space: O(1)

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