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.
Output: 3 5 12 12
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 2j-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 2j-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.
Please Login to comment...