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)

use std::io::*;
use std::str::FromStr;

type NodeId = usize;

struct Node {
    parent: Option<NodeId>,
    left_child: Option<NodeId>,
    right_brother: Option<NodeId>,
    depth: u32,
}

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

fn rec(node_id: NodeId, depth: u32, nodes: &mut Vec<Node>) {
    nodes[node_id].depth = depth;
    if nodes[node_id].right_brother != None {
        rec(nodes[node_id].right_brother.unwrap(), depth, nodes)
    }
    if nodes[node_id].left_child != None {
        rec(nodes[node_id].left_child.unwrap(), depth + 1, nodes)
    }
}

fn main() {
    let n: usize = read();
    let mut nodes: Vec<Node> = (0..n)
        .map(|_| Node {
            parent: None,
            left_child: None,
            right_brother: None,
            depth: 0,
        })
        .collect();

    let mut root_node_id: NodeId = 0;

    for _ in 0..n {
        let node_id: NodeId = read();
        let k: usize = read();
        let mut left_brother_id: Option<NodeId> = None;
        for index in 0..k {
            let child_node_id: NodeId = read();
            if index == 0 {
                nodes[node_id].left_child = Some(child_node_id);
            } else {
                nodes[left_brother_id.unwrap()].right_brother = Some(child_node_id);
            }
            left_brother_id = Some(child_node_id);
            nodes[child_node_id].parent = Some(node_id);
        }
    }

    for index in 0..n {
        if nodes[index].parent == None {
            root_node_id = index;
        }
    }

    rec(root_node_id, 0, &mut nodes);

    for (index, node) in nodes.iter().enumerate() {
        print!("node {}: ", index);
        if node.parent == None {
            print!("parent = -1, ");
        } else {
            print!("parent = {}, ", node.parent.unwrap());
        }
        print!("depth = {}, ", node.depth);

        if node.parent == None {
            print!("root, ");
        } else if node.left_child == None {
            print!("leaf, ");
        } else {
            print!("internal node, ");
        }

        print!("[");
        if node.left_child != None {
            let mut count = 0;
            let mut right_brother_id: Option<NodeId> = node.left_child;
            while right_brother_id != None {
                let right_brother: &Node = &nodes[right_brother_id.unwrap()];
                if count != 0 {
                    print!(", ")
                }
                print!("{}", right_brother_id.unwrap());
                count += 1;
                right_brother_id = right_brother.right_brother
            }
        }
        println!("]");
    }
}