0
$\begingroup$

Given a tree of "N" nodes(each node has been assigned a value A[i],node-"1" is the root of the tree), and a constant "K" , we have Q queries of the following type : [w]

(which means find the lowest valued node in the sub-tree of [w] , only considering those nodes in the sub-tree of [w] which have a depth less than equal to K) .

Example :

Value of nodes of tree :

A[1] = 10

A[2] = 20

A[3] = 30

A[4] = 40

A[5] = 50

A[6] = 60

Edges of tree : [1-2],

[2-3],

[3-4],

[4-5],

[4-6].

K=2.

Query-1 : [w]=1 . All nodes in subtree of [w] : (1,2,3,4,5,6) , now, all nodes in sub-tree of [w] having depth less than equal to K : (1,2) . Hence , minimum(A[1],A[2])=min(10,20)=10 is the answer .

Query-2 : [w]=4 . All nodes in subtree of [w] : (4,5,6) , now, all nodes in sub-tree of [w] having depth less than equal to K : (4,5,6). Hence , minimum(A[4],A[5],A[6]) = min(40,50,60)=40 is the answer .

$\endgroup$
1
  • $\begingroup$ Why not just precompute the answer for all nodes? Can you please also explain what you've tried and where the problem comes from? $\endgroup$
    – user114966
    Commented Jul 27, 2020 at 4:41

1 Answer 1

1
$\begingroup$

Without any precomputation, a depth-limited version of depth-first search would be a natural approach to this problem: then it's like finding the minimum of an array, you could find the minimum weight during the search. But a time-efficient way (if you have multiple queries for different nodes and $K$) is to indeed precompute the answer for all the nodes: a linear algorithm seems achievable, starting from the bottom level up to the root, level by level. You have to store for each node an array $a$ such as $a(K)$ gives the desired value. And if you modify the value of a node later, I think a backpropagation scheme could be also considered.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.