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 Go (judge verdict: Accepted)

package main

import (
	"bufio"
	"bytes"
	"fmt"
	"os"
	"strconv"
)

func main() {
	scn := bufio.NewScanner(os.Stdin)
	scn.Split(bufio.ScanWords)
	nextInt := func() int { scn.Scan(); r, _ := strconv.Atoi(scn.Text()); return r }

	n := nextInt()

	parents := make([]int, n)
	for i := range parents {
		parents[i] = -1
	}
	children := make([][]int, n)

	for i := 0; i < n; i++ {
		id, k := nextInt(), nextInt()
		children[id] = make([]int, 0, k)
		for j := 0; j < k; j++ {
			c := nextInt()
			parents[c] = id
			children[id] = append(children[id], c)
		}
	}

	root := -1
	for id, parent := range parents {
		if parent == -1 {
			root = id
		}
	}

	depths := make([]int, n)
	markDepth(depths, 1, children, root)

	for id := 0; id < n; id++ {
		typ := "internal node"
		if len(children[id]) == 0 {
			typ = "leaf"
		}
		if depths[id] == 0 {
			typ = "root"
		}
		fmt.Printf("node %d: parent = %d, depth = %d, %s, %s\n", id, parents[id], depths[id], typ, formatArr(children[id]))
	}
}

func markDepth(depths []int, depth int, children [][]int, root int) {
	for _, c := range children[root] {
		depths[c] = depth
		markDepth(depths, depth+1, children, c)
	}
}

func formatArr(a []int) string {
	w := bytes.NewBuffer(nil)
	w.WriteString("[")
	for i, n := range a {
		if i > 0 {
			fmt.Fprint(w, ", ")
		}
		fmt.Fprint(w, n)
	}
	w.WriteString("]")
	return w.String()
}