Find distance between two nodes in the given Binary tree for Q queries

Given a binary tree having N nodes and weight of N-1 edges. The distance between two nodes is the sum of the weight of edges on the path between two nodes. Each query contains two integers U and V, the task is to find the distance between nodes U and V.

Examples: 

Input: 
 

👁 Image

Output: 3 5 12 12 
Explanation: 
Distance between nodes 1 to 3 = weight(1, 3) = 2 
Distance between nodes 2 to 3 = weight(1, 2) + weight(1, 3) = 5 
Distance between nodes 3 to 5 = weight(1, 3) + weight(1, 2) + weight(2, 5) = 12 
Distance between nodes 4 to 5 = weight(4, 2) + weight(2, 5) = 12 
 

Approach: The idea is to use LCA in a tree using Binary Lifting Technique.  

• Binary Lifting is a Dynamic Programming approach where we pre-compute an array lca[i][j] where i = [1, n], j = [1, log(n)] and lca[i][j] contains 2^j-th ancestor of node i. 
  
  • For computing the values of lca[][], the following recursion may be used

• As we will compute the lca[][] array we will also calculate the distance[][] where distance[i][j] contains the distance from node i to its 2^j-th ancestor 
  
  • For computing the values of dist[][], the following recursion may be used.

• After precomputation, we find the distance between (u, v) as we find the least common ancestor of (u, v).

Below is the implementation of the above approach:

3
5
12

Time Complexity: The time taken in pre-processing is O(N logN) and every query takes O(logN) time. Therefore, overall time complexity of the solution is O(N logN).

Space Complexity: O(N*log(N)) 
We are storing the LCA and distance of all the nodes in two 2-D arrays.