Problem:
Rooted Trees

A graph G = (V, E) is a data structure where V is a finite set of vertices and E is a binary relation on V represented by a set of edges. Fig. 1 illustrates an example of a graph (or graphs).

Fig. 1

A free tree is a connnected, acyclic, undirected graph. A rooted tree is a free tree in which one of the vertices is distinguished from the others. A vertex of a rooted tree is called "node."

Your task is to write a program which reports the following information for each node u of a given rooted tree T:

node ID of u

parent of u

depth of u

node type (root, internal node or leaf)

a list of chidlren of u

If the last edge on the path from the root r of a tree T to a node x is (p, x), then p is the parent of x, and x is a child of p. The root is the only node in T with no parent.

A node with no children is an external node or leaf. A nonleaf node is an internal node

The number of children of a node x in a rooted tree T is called the degree of x.

The length of the path from the root r to a node x is the depth of x in T.

Here, the given tree consists of n nodes and evey node has a unique ID from 0 to n-1.

Fig. 2 shows an example of rooted trees where ID of each node is indicated by a number in a circle (node). The example corresponds to the first sample input.

Fig. 2

Input

The first line of the input includes an integer n, the number of nodes of the tree.

In the next n lines, the information of each node u is given in the following format:

id k c1 c2 ... ck

where id is the node ID of u, k is the degree of u, c1 ... ck are node IDs of 1st, ... kth child of u. If the node does not have a child, the k is 0.

Output

Print the information of each node in the following format ordered by IDs:

node id: parent = p, depth = d, type, [c1...ck]

p is ID of its parent. If the node does not have a parent, print -1.

d is depth of the node.

type is a type of nodes represented by a string (root, internal node or leaf). If the root can be considered as a leaf or an internal node, print root.

c1...ck is the list of children as a ordered tree.

Please follow the format presented in a sample output below.

Constraints

1 ≤ n ≤ 100000

Sample Input 1

13
0 3 1 4 10
1 2 2 3
2 0
3 0
4 3 5 6 7
5 0
6 0
7 2 8 9
8 0
9 0
10 2 11 12
11 0
12 0

Sample Output 1

node 0: parent = -1, depth = 0, root, [1, 4, 10]
node 1: parent = 0, depth = 1, internal node, [2, 3]
node 2: parent = 1, depth = 2, leaf, []
node 3: parent = 1, depth = 2, leaf, []
node 4: parent = 0, depth = 1, internal node, [5, 6, 7]
node 5: parent = 4, depth = 2, leaf, []
node 6: parent = 4, depth = 2, leaf, []
node 7: parent = 4, depth = 2, internal node, [8, 9]
node 8: parent = 7, depth = 3, leaf, []
node 9: parent = 7, depth = 3, leaf, []
node 10: parent = 0, depth = 1, internal node, [11, 12]
node 11: parent = 10, depth = 2, leaf, []
node 12: parent = 10, depth = 2, leaf, []

Sample Input 2

4
1 3 3 2 0
0 0
3 0
2 0

Sample Output 2

node 0: parent = 1, depth = 1, leaf, []
node 1: parent = -1, depth = 0, root, [3, 2, 0]
node 2: parent = 1, depth = 1, leaf, []
node 3: parent = 1, depth = 1, leaf, []

Note

You can use a left-child, right-sibling representation to implement a tree which has the following data:

the parent of u

the leftmost child of u

the immediate right sibling of u

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in Rust (judge verdict: Accepted)

#[allow(unused_imports)]
use std::cmp::{max, min};
use std::fmt;
#[allow(unused_imports)]
use std::io::{stdin, stdout, Read, Write};
use std::str::FromStr;

#[derive(Default)]
struct Scanner<R: Read> {
    reader: R,
}

impl<R: Read> Scanner<R> {
    fn new(reader: R) -> Scanner<R> {
        Scanner { reader }
    }

    fn scan<T: FromStr>(&mut self) -> Option<T> {
        let s = self
            .reader
            .by_ref()
            .bytes()
            .map(|c| c.unwrap() as char)
            .skip_while(|c| c.is_whitespace())
            .take_while(|c| !c.is_whitespace())
            .collect::<String>();
        s.parse::<T>().ok()
    }

    fn next<T: FromStr>(&mut self) -> T {
        if let Some(s) = self.scan() {
            s
        } else {
            std::process::exit(1);
        }
    }
}

#[derive(Default)]
struct Node {
    id: usize,
    parent: usize,
    depth: usize,
    children: Vec<usize>,
}

impl fmt::Display for Node {
    fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
        let t = if self.depth == 0 {
            "root"
        } else if self.children.len() == 0 {
            "leaf"
        } else {
            "internal node"
        };

        let mut p = self.parent as i32;
        if self.parent == self.id {
            p = -1;
        }

        write!(
            f,
            "node {}: parent = {}, depth = {}, {}, {:?}",
            self.id, p, self.depth, t, self.children
        )
    }
}

/*
fn dfs(i: usize, d: usize, n: &mut [Node]) {
    n[i].depth = d;
    for &j in n[i].children.iter() {
        dfs(j, d+1, n);
    }
}
*/
fn main() {
    let cin = stdin();
    let cin = cin.lock();
    let mut sc = Scanner::new(cin);
    let out = stdout();
    let mut out = out.lock();

    let n = sc.next();
    let mut node = Vec::<Node>::new();
    for i in 0..n {
        node.push(Node {
            id: i,
            parent: i,
            depth: 0,
            children: Vec::new(),
        })
    }
    for _ in 0..n {
        let id: usize = sc.next();
        let k = sc.next();
        for _ in 0..k {
            let c = sc.next();
            node[id].children.push(c);
            node[c].parent = id;
        }
    }

    let mut root = 0;
    for x in node.iter() {
        if x.parent == x.id {
            root = x.id;
            break;
        }
    }

    let mut stack = vec![(root, 0)];
    while !stack.is_empty() {
        let (i, d) = stack.pop().unwrap();
        node[i].depth = d;
        for &c in node[i].children.iter() {
            stack.push((c, d+1));
        }
    }

    for x in node.iter() {
        writeln!(out, "{}", x).ok();
    }
}