#####   ######  #####    ###    #   #  ###  #   #  ######
##  ##  ##      ##  ##  ## ##   #   #   #   #   #  ##
#####   ####    #####   #   #   #   #   #   #   #  ####
##  #   ##      ##      ## ##    # #    #    # #   ##
##   #  ######  ##       ###      #    ###    #    ######

$ curl repovive.com/roadmaps/graph-theory/mixed-practice-tree-fundamentals/tree-distances-ii-implementation

░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░███████████████████████████████████████████████████████████████████████████████████████████████████████████

#####   ######  #####    ###    #   #  ###  #   #  ######
##  ##  ##      ##  ##  ## ##   #   #   #   #   #  ##
#####   ####    #####   #   #   #   #   #   #   #  ####
##  #   ##      ##      ## ##    # #    #    # #   ##
##   #  ######  ##       ###      #    ###    #    ######

$ curl repovive.com/roadmaps/graph-theory/mixed-practice-tree-fundamentals/tree-distances-ii-implementation

Repovive

Tree Distances II - Implementation - Mixed Practice: Tree Fundamentals | Graph Theory | Repovive

Repovive.

Graph Theory37 sections · 1633 units81h total

Open in Course

Tree Distances II - Implementation

(Rerooting DP)

Here is the solution:

sub = array of size n
down = array of size n
ans = array of size n

function dfs1(u, p):
    sub[u] = 1
    for v in adj[u]:
        if v == p:
            continue
        dfs1(v, u)
        sub[u] = sub[u] + sub[v]
        down[u] = down[u] + down[v] + sub[v]

function dfs2(u, p, up):
    ans[u] = down[u] + up
    for v in adj[u]:
        if v == p:
            continue
        dfs2(v, u, up + (down[u] - down[v] - sub[v]) + (n - sub[v]))

This runs in $O(n)$ time and uses $O(n)$ space.