A14785. Distance Tree (hard version)

This version of the problem differs from the previous one only in the constraint on $n$ .

A tree is a connected undirected graph without cycles. A weighted tree has a weight assigned to each edge. The distance between two vertices is the minimum sum of weights on the path connecting them.

You are given a weighted tree with $n$ vertices, each edge has a weight of $1$ . Denote $d(v)$ as the distance between vertex $1$ and vertex $v$ .

Let $f(x)$ be the minimum possible value of $\max\limits_{1 \leq v \leq n} \ {d(v)}$ if you can temporarily add an edge with weight $x$ between any two vertices $a$ and $b$ $(1 \le a, b \le n)$ . Note that after this operation, the graph is no longer a tree.

For each integer $x$ from $1$ to $n$ , find $f(x)$ .

The first line contains a single integer $t$ ( $1 \le t \le 10^4$ ) — the number of test cases.

The first line of each test case contains a single integer $n$ ( $2 \le n \le 3 \cdot 10^5$ ).

Each of the next $n−1$ lines contains two integers $u$ and $v$ ( $1 \le u,v \le n$ ) indicating that there is an edge between vertices $u$ and $v$ . It is guaranteed that the given edges form a tree.

It is guaranteed that the sum of $n$ over all test cases doesn't exceed $3 \cdot 10^5$ .

For each test case, print $n$ integers in a single line, $x$ -th of which is equal to $f(x)$ for all $x$ from $1$ to $n$ .

 In the first testcase: - For $x = 1$ , we can an edge between vertices $1$ and $3$ , then $d(1) = 0$ and $d(2) = d(3) = d(4) = 1$ , so $f(1) = 1$ .
- For $x \ge 2$ , no matter which edge we add, $d(1) = 0$ , $d(2) = d(4) = 1$ and $d(3) = 2$ , so $f(x) = 2$ .