LCA in a tree using Binary Lifting Technique
Given a binary tree, the task is to find the Lowest Common Ancestor of the given two nodes in the tree.
Let G be a tree then the LCA of two nodes u and v is defined as the node w in the tree which is an ancestor of both u and v and is farthest from the root node. If one node is the ancestor of another one then that particular node is the LCA of those two nodes. In particular, if we want to find the LCA of u and v, and if u is a parent of v, then u is their lowest common ancestor.
The LCA of 6 and 9 is 1.
The LCA of 5 and 9 is 1.
The LCA of 6 and 8 is 3.
The LCA of 6 and 1 is 1.
Approach: The article describes an approach known as Binary Lifting to find the Lowest Common Ancestor of two nodes in a tree. There can be many approaches to solve the LCA problem. We are discussing the Binary Lifting Technique, the others can be read from here and here.
Binary Lifting is a dynamic programming approach where we pre-compute an array memo[1, n][1, log(n)] where memo[i][j] contains 2^j-th ancestor of node i. For computing the values of memo, the following recursion may be used:
memo state: memo[i][j] = i-th node's 2^(j)th ancestor in the path memo initialization: memo[i][j] = memo[i] (first parent (2^0) of each node is given) memo trans: memo[i][j] = memo[ memo [i][j - 1]] meaning: A(i,2^j)=A( A(i , 2^(j-1) ) , 2^(j-1) ) To find the (2^j)-th ancestor of i, recursively find i-th node's 2^(j-1)th ancestor's 2^(j-1)th ancestor. (2^(j) = 2^(j-1) + 2^(j-1)) So: memo[i][j] = parent[i] if j = 0 and memo[i][j] = memo[memo[i][j - 1]][j - 1] if j > 0.
We first check whether a node is an ancestor of another or not and if one node is ancestor of another then it is the LCA of these two nodes otherwise we find a node that is not the common ancestor of both u and v and is the highest(i.e. a node x such that x is not the common ancestor of u and v but memo[x] is) in the tree. After finding such a node (let it be x), we print the first ancestor of x i.e. memo[x] which will be the required LCA.
Below is the implementation of the above approach:
The LCA of 6 and 9 is 1 The LCA of 5 and 9 is 1 The LCA of 6 and 8 is 3 The LCA of 6 and 1 is 1
Time Complexity: The time taken in pre-processing is O(NlogN) and every query takes O(logN) time. So the overall time complexity of the solution is O(NlogN)
Auxiliary Space: O(NlogN)
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