# Tree Coordinates

### Problem Statement :

We consider metric space to be a pair, , where  is a set and  such that the following conditions hold:

where  is the distance between points  and .

Let's define the product of two metric spaces, , to be  such that:

, where , .
So, it follows logically that  is also a metric space. We then define squared metric space, , to be the product of a metric space multiplied with itself: .

For example, , where  is a metric space. , where .

In this challenge, we need a tree-space. You're given a tree, , where  is the set of vertices and  is the set of edges. Let the function  be the distance between two vertices in tree  (i.e.,  is the number of edges on the path between vertices  and ). Note that  is a metric space.

You are given a tree, , with  vertices, as well as  points in . Find and print the distance between the two furthest points in this metric space!

Input Format

The first line contains two space-separated positive integers describing the respective values of  (the number of vertices in ) and  (the number of given points).
Each line  of the  subsequent lines contains two space-separated integers,  and , describing edge  in .
Each line  of the  subsequent lines contains two space-separated integers describing the respective values of  and  for point .

Constraints

Scoring

This challenge uses binary scoring, so you must pass all test cases to earn a positive score.

Output Format

Print a single non-negative integer denoting the maximum distance between two of the given points in metric space .

### Solution :

Solution in C :

In C ++ :

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <cstdlib>
#include <ctime>
#include <deque>
using namespace std;

#define N 200000

int fa[N][21], dep[N], L[N], R[N], Time, n, m;
vector <int> ve[N], ve1[N];
bool used[N];
int zx, sz;
int sum[N], l[N], c[N];
int ans;
int ma1[N], ma2[N], c1[N], c2[N];
int zhan[N], zhan1[N];

void dfs1(int k, int f) {
fa[k][0] = f;
dep[k] = dep[f] + 1;
for (int i = 1; i <= 20; i++)
fa[k][i] = fa[fa[k][i - 1]][i - 1];
L[k] = ++Time;
for (int i = 0; i < (int) ve[k].size(); i++)
if (ve[k][i] != f)
dfs1(ve[k][i], k);
R[k] = Time;
}

void dfs_size(int k) {
sum[k] = 1;
used[k] = true;
for (int i = 0; i < (int) ve[k].size(); i++)
if (!used[ve[k][i]]) {
dfs_size(ve[k][i]);
sum[k] += sum[ve[k][i]];
}
used[k] = false;
}

void dfs_zx(int k) {
// printf("?? %d %d\n", k, sz);
bool ok = (sz - sum[k] <= sz / 2);
used[k] = true;
for (int i = 0; i < (int) ve[k].size(); i++)
if (!used[ve[k][i]]) {
if (sum[ve[k][i]] > sz / 2)
ok = false;
dfs_zx(ve[k][i]);
}
used[k] = false;
if (ok)
zx = k;
}

void dfs_l(int k) {
used[k] = true;
for (int i = 0; i < (int) ve[k].size(); i++)
if (!used[ve[k][i]]) {
l[ve[k][i]] = l[k] + 1;
dfs_l(ve[k][i]);
}
used[k] = false;
}

void dfs_c(int k) {
used[k] = true;
for (int i = 0; i < (int) ve[k].size(); i++)
if (!used[ve[k][i]]) {
c[ve[k][i]] = c[k];
dfs_c(ve[k][i]);
}
used[k] = false;
}

bool cmp_c(pair<int, int> x, pair<int, int> y) {
return c[x.first] < c[y.first];
}

bool cmp_L(int x, int y) {
return L[x] < L[y];
}

int dist(int x, int y) {
int ans = 0;
if (dep[x] < dep[y])
swap(x, y);
for (int i = 20; i >= 0; i--)
if (dep[fa[x][i]] >= dep[y])
ans += (1 << i), x = fa[x][i];
if (x == y)
return ans;
for (int i = 20; i >= 0; i--)
if (fa[x][i] != fa[y][i])
ans += 1 << (i + 1), x = fa[x][i], y = fa[y][i];
return ans + 2;
}

int lca(int x, int y) {
int ans = 0;
if (dep[x] < dep[y])
swap(x, y);
for (int i = 20; i >= 0; i--)
if (dep[fa[x][i]] >= dep[y])
ans += (1 << i), x = fa[x][i];
if (x == y)
return x;
for (int i = 20; i >= 0; i--)
if (fa[x][i] != fa[y][i])
ans += 1 << (i + 1), x = fa[x][i], y = fa[y][i];
return fa[x][0];
}

void dfs_d(int k) {
// ma1[k] = -1000000000;
// ma2[k] = -1000000000;
// ma1[k] = v[k];
// c1[k] = c[k];
// printf("?? %d\n", k);
for (int i = 0; i < (int) ve1[k].size(); i++) {
int t = ve1[k][i];
dfs_d(t);
ma1[t] += dep[t] - dep[k];
ma2[t] += dep[t] - dep[k];
if (c1[k] != c1[t] || !c1[k] || !c1[t])
ans = max(ans, ma1[k] + ma1[t]);
if (c1[k] != c2[t] || !c1[k] || !c2[t])
ans = max(ans, ma1[k] + ma2[t]);

if (c2[k] != c1[t] || !c2[k] || !c1[t])
ans = max(ans, ma2[k] + ma1[t]);

if (c2[k] != c2[t] || !c2[k] || !c2[t])
ans = max(ans, ma2[k] + ma2[t]);

if (ma1[t] > ma1[k]) {
if (c1[t] == c1[k])
ma1[k] = ma1[t];
else {
ma2[k] = ma1[k];
c2[k] = c1[k];
ma1[k] = ma1[t];
c1[k] = c1[t];
}
}else if (ma1[t] > ma2[k] && c1[t] != c1[k]) {
ma2[k] = ma1[t];
c2[k] = c1[t];
}
if (ma2[t] > ma1[k]) {
if (c2[t] == c1[k])
ma1[k] = ma2[t];
else {
ma2[k] = ma1[k];
c2[k] = c1[k];
ma1[k] = ma2[t];
c1[k] = c2[t];
}
}else if (ma2[t] > ma2[k] && c2[t] != c1[k]) {
ma2[k] = ma2[t];
c2[k] = c2[t];
}

}
}

void doit(vector <pair<int, int> > V) {
int len = 0;
// printf("%d\n", ans);
for (int i = 0; i < (int) V.size(); i++) {
zhan[++len] = V[i].second;

if (c[V[i].first] != c1[V[i].second] || !c[V[i].first] || !c1[V[i].second]) {
ans = max(ans, ma1[V[i].second] + l[V[i].first]);
}
if (c[V[i].first] != c2[V[i].second] || !c[V[i].first] || !c2[V[i].second]) {
ans = max(ans, ma2[V[i].second] + l[V[i].first]);
}

if (l[V[i].first] > ma1[V[i].second]) {
if (c[V[i].first] != c1[V[i].second]) {
ma2[V[i].second] = ma1[V[i].second];
c2[V[i].second] = c1[V[i].second];
ma1[V[i].second] = l[V[i].first];
c1[V[i].second] = c[V[i].first];
}else {
ma1[V[i].second] = l[V[i].first];
}
}else if (l[V[i].first] > ma2[V[i].second] && c[V[i].first] != c1[V[i].second]) {
ma2[V[i].second] = l[V[i].first];
c2[V[i].second] = c[V[i].first];
}

}
// printf("%d\n", ans);
sort(zhan + 1, zhan + len + 1, cmp_L);

for (int i = len; i > 1; i--) {
zhan[++len] = lca(zhan[i - 1], zhan[i]);
}
sort(zhan + 1, zhan + len + 1, cmp_L);

for (int i = 1; i < len; i++) {
ve1[zhan[i]].clear();
}
// for (int i = 1; i <= len; i++)
// 	printf("%d ", zhan[i]);
// printf("\n");
int len1 = 0;
zhan1[len1 = 1] = zhan[1];
for (int i = 2; i <= len; i++)
if (zhan[i] != zhan[i - 1]) {
while (len1 > 0 && R[zhan1[len1]] < L[zhan[i]])
len1 --;
zhan1[++len1] = zhan[i];
ve1[zhan1[len1 - 1]].push_back(zhan1[len1]);
}
dfs_d(zhan[1]);
// for (int i = 0; i < (int) V.size(); i++)
// 	for (int j = 0; j < i; j++)
// 		if (c[V[i].first] != c[V[j].first] || !c[V[i].first] || !c[V[j].first]) {
// 			// printf("xx %d %d\n", l[V[i].first] + l[V[j].first], dist(V[i].second, V[j].second));
// 			ans = max(ans, l[V[i].first] + l[V[j].first] + dist(V[i].second, V[j].second));
// 		}
for (int i = 1; i <= len; i++) {
// v[V[i].second] = -1000000000;
ma1[zhan[i]] = ma2[zhan[i]] = -1000000000;
}
}

void doit1(vector <pair<int, int> > V) {
for (int i = 0; i < (int) V.size(); i++)
for (int j = 0; j < i; j++)
// if (c[V[i].first] != c[V[j].first] || !c[V[i].first] || !c[V[j].first]) {
// printf("xx %d %d\n", l[V[i].first] + l[V[j].first], dist(V[i].second, V[j].second));
ans = max(ans, dist(V[i].first, V[j].first) + dist(V[i].second, V[j].second));
// }
}

void dfs(int k, vector <pair<int, int> > V) {
// printf("?? %d\n", k);
if (V.size() <= 1)
return ;
if (V.size() <= 100) {
doit1(V);
return ;
}
dfs_size(k);
sz = sum[k];
dfs_zx(k);
k = zx;
l[k] = 0;
dfs_l(k);
c[k] = 0;
used[k] = true;
int tot = 0;
for (int i = 0; i < (int) ve[k].size(); i++)
if (!used[ve[k][i]]) {
tot += 1;
c[ve[k][i]] = tot;
dfs_c(ve[k][i]);
}
used[k] = false;

doit(V);

sort(V.begin(), V.end(), cmp_c);
int q = 0;
used[k] = true;
vector <int> vv;
for (int i = 0; i <= tot; i++) {
while (q < (int) V.size() && c[V[q].first] <= i)
q += 1;
vv.push_back(q);
}
tot = 0;
for (int i = 0; i < (int) ve[k].size(); i++)
if (!used[ve[k][i]]) {
tot += 1;
vector <pair<int, int> > V1;
// printf("?? %d %d %u\n", vv[tot - 1], vv[tot], V.size());
for (int j = vv[tot - 1]; j < vv[tot]; j++)
V1.push_back(V[j]);
dfs(ve[k][i], V1);
}
used[k] = false;
}

int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i < n; i++) {
int x, y;
scanf("%d%d", &x, &y);
// x = i;
// y = i + 1;
ve[x].push_back(y);
ve[y].push_back(x);
}
vector <pair<int, int> > V;
for (int i = 1; i <= m; i++) {
int x, y;
scanf("%d%d", &x, &y);
V.push_back(make_pair(x, y));
}
dfs1(1, 0);
for (int i = 1; i <= n; i++)
ma1[i] = ma2[i] = -1000000000;
dfs(1, V);
printf("%d\n", ans);
}

In Java :

import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;
import java.util.regex.*;

public class Solution {

{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;

{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
}

{
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
}

{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1)
{
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}

public int nextInt() throws IOException
{
int ret = 0;
while (c <= ' ')
boolean neg = (c == '-');
if (neg)
do
{
ret = ret * 10 + c - '0';
}  while ((c = read()) >= '0' && c <= '9');

if (neg)
return -ret;
return ret;
}

public long nextLong() throws IOException
{
long ret = 0;
while (c <= ' ')
boolean neg = (c == '-');
if (neg)
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}

public double nextDouble() throws IOException
{
double ret = 0, div = 1;
while (c <= ' ')
boolean neg = (c == '-');
if (neg)

do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');

if (c == '.')
{
while ((c = read()) >= '0' && c <= '9')
{
ret += (c - '0') / (div *= 10);
}
}

if (neg)
return -ret;
return ret;
}

private void fillBuffer() throws IOException
{
buffer[0] = -1;
}

{
fillBuffer();
return buffer[bufferPointer++];
}

public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}

static int[][] buildSparseTable(int[] arr) {
int pow = 1;
while ((1 << pow) < arr.length) pow++;
int[][] result = new int[arr.length][pow];
for (int i = 0; i < arr.length; i++) result[i][0] = arr[i];
for (int j = 1; j <= pow; j++) {
for (int i = 0; i + (1 << j) <= arr.length; i++) {
result[i][j] = Math.min(result[i][j-1],
result[i + (1 << (j-1))][j-1]);
}
}
return result;
}

/*
* Complete the treeCoordinates function below.
*/
static int treeCoordinates(int n, int[][] edges, int[][] points) {
ArrayList<Integer>[] nodes = new ArrayList[n + 1];
for (int i = 1; i <= n; i++) nodes[i] = new ArrayList<Integer>();
for (int[] edge : edges) {
}

// Find diameter (two BFS)
int root = 0;
int tail = 0;
{
class Entry {
int node;
int dist;
public Entry(int node, int dist) {
this.node = node;
this.dist = dist;
}
}
boolean[] visited = new boolean[n + 1];
visited[1] = true;
Q.offer(new Entry(1, 0));
int maxDist = 0;
int farNode = 1;
while (Q.size() > 0) {
Entry cur = Q.poll();
if (cur.dist > maxDist) {
maxDist = cur.dist;
farNode = cur.node;
}
for (int neighbor : nodes[cur.node]) {
if (visited[neighbor]) continue;
visited[neighbor] = true;
Q.offer(new Entry(neighbor, cur.dist + 1));
}
}
root = farNode;

visited = new boolean[n + 1];
visited[root] = true;
Q.offer(new Entry(root, 0));
maxDist = 0;
farNode = root;
while (Q.size() > 0) {
Entry cur = Q.poll();
if (cur.dist > maxDist) {
maxDist = cur.dist;
farNode = cur.node;
}
for (int neighbor : nodes[cur.node]) {
if (visited[neighbor]) continue;
visited[neighbor] = true;
Q.offer(new Entry(neighbor, cur.dist + 1));
}
}
tail = farNode;
}
//System.out.println("root = " + root + ", tail = " + tail);

// Euler tour
int[] eulerTour = new int[2*n - 1];
final int[] depth = new int[n + 1];
int[] eulerLevels = new int[2*n - 1];
int[] eulerIndex = new int[n + 1];
boolean[] visited = new boolean[n + 1];

int[] S = new int[n];
int spos = 0;
S[0] = root;
int pos = 0;
int[] neighborCount = new int[n + 1];
while (spos > -1) {
int cur = S[spos--];
if (!visited[cur]) {
depth[cur] = spos + 1;
eulerIndex[cur] = pos;
visited[cur] = true;
}
eulerLevels[pos] = spos + 1;
eulerTour[pos] = cur;
pos++;
while (neighborCount[cur] < nodes[cur].size()) {
if (visited[nodes[cur].get(neighborCount[cur])]) {
neighborCount[cur]++;
continue;
}
int next = nodes[cur].get(neighborCount[cur]);
//parent[next] = cur;
S[++spos] = cur;
S[++spos] = next;
neighborCount[cur]++;
break;
}
}
//System.out.println(Arrays.toString(eulerTour));
//System.out.println(Arrays.toString(eulerLevels));
//System.out.println(Arrays.toString(eulerIndex));

// Sparse table
int[][] lookup = buildSparseTable(eulerLevels);
//for (int i = 0; i < lookup.length; i++) {
//    for (int j = 0; j < lookup[0].length; j++) {
//        System.out.print(lookup[i][j] + " ");
//    }
//    System.out.println();
//}

class Entry implements Comparable<Entry> {
int x;
int y;
int val;
public Entry(int x, int y, int val) {
this.x = x;
this.y = y;
this.val = val;
}
public int compareTo(Entry other) {
return val - other.val;
}
}
List<Entry> list1 = new ArrayList<Entry>(points.length);
List<Entry> list2 = new ArrayList<Entry>(points.length);
List<Entry> list3 = new ArrayList<Entry>(points.length);
List<Entry> list4 = new ArrayList<Entry>(points.length);

for (int i = 0; i < points.length; i++) {
int x = points[i][0];
int y = points[i][1];
int xLcaLevel;
{
int start = Math.min(eulerIndex[x], eulerIndex[tail]);
int end = Math.max(eulerIndex[x], eulerIndex[tail]);
int pow = 0;
while (1 << (pow + 1) <= (end - start)) pow++;
xLcaLevel = Math.min(lookup[start][pow],
lookup[end + 1 - (1<<pow)][pow]);
}
int yLcaLevel;
{
int start = Math.min(eulerIndex[y], eulerIndex[tail]);
int end = Math.max(eulerIndex[y], eulerIndex[tail]);
int pow = 0;
while (1 << (pow + 1) <= (end - start)) pow++;
yLcaLevel = Math.min(lookup[start][pow],
lookup[end + 1 - (1<<pow)][pow]);
}
int val1 = depth[x] + depth[y];
int val2 = -depth[x] - depth[y] + 2*xLcaLevel + 2*yLcaLevel;
int val3 = depth[x] + depth[y] - 2*xLcaLevel;
int val4 = -depth[x] - depth[y] + 2*yLcaLevel;
}
Collections.sort(list1, Collections.reverseOrder());
Collections.sort(list2);
Collections.sort(list3, Collections.reverseOrder());
Collections.sort(list4);

//int maxDist = Math.max(list1.get(0).val - list2.get(0).val, list3.get(0).val - list4.get(0).val);
int maxDist = 0;

for (int i = 0; i < points.length; i++) {
// ith increasing diagonal
boolean shouldContinue = false;
for (int j = 0; j <= i; j++) {
Entry e1 = list1.get(i-j);
Entry e2 = list2.get(j);
int potential12 = e1.val - e2.val;
if (potential12 > maxDist) {
shouldContinue = true;
int x1 = e1.x;
int y1 = e1.y;
int x2 = e2.x;
int y2 = e2.y;
int xLcaLevel;
{
int start = Math.min(eulerIndex[x1], eulerIndex[x2]);
int end = Math.max(eulerIndex[x1], eulerIndex[x2]);
int pow = 0;
while (1 << (pow + 1) <= (end - start)) pow++;
xLcaLevel = Math.min(lookup[start][pow],
lookup[end + 1 - (1<<pow)][pow]);
}
int yLcaLevel;
{
int start = Math.min(eulerIndex[y1], eulerIndex[y2]);
int end = Math.max(eulerIndex[y1], eulerIndex[y2]);
int pow = 0;
while (1 << (pow + 1) <= (end - start)) pow++;
yLcaLevel = Math.min(lookup[start][pow],
lookup[end + 1 - (1<<pow)][pow]);
}
int actual12 = depth[x1] + depth[x2] - 2*xLcaLevel
+ depth[y1] + depth[y2] - 2*yLcaLevel;
maxDist = Math.max(maxDist, actual12);
}
Entry e3 = list3.get(i-j);
Entry e4 = list4.get(j);
int potential34 = e3.val - e4.val;
if (potential34 > maxDist) {
shouldContinue = true;
int x3 = e3.x;
int y3 = e3.y;
int x4 = e4.x;
int y4 = e4.y;
int xLcaLevel;
{
int start = Math.min(eulerIndex[x3], eulerIndex[x4]);
int end = Math.max(eulerIndex[x3], eulerIndex[x4]);
int pow = 0;
while (1 << (pow + 1) <= (end - start)) pow++;
xLcaLevel = Math.min(lookup[start][pow],
lookup[end + 1 - (1<<pow)][pow]);
}
int yLcaLevel;
{
int start = Math.min(eulerIndex[y3], eulerIndex[y4]);
int end = Math.max(eulerIndex[y3], eulerIndex[y4]);
int pow = 0;
while (1 << (pow + 1) <= (end - start)) pow++;
yLcaLevel = Math.min(lookup[start][pow],
lookup[end + 1 - (1<<pow)][pow]);
}
int actual34 = depth[x3] + depth[x4] - 2*xLcaLevel
+ depth[y3] + depth[y4] - 2*yLcaLevel;
maxDist = Math.max(maxDist, actual34);
}
}
if (!shouldContinue) break;
}

return maxDist;
}

public static void main(String[] args) throws IOException {
BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

int[][] edges = new int[n-1][2];

for (int edgesRowItr = 0; edgesRowItr < n-1; edgesRowItr++) {
}

int[][] points = new int[m][2];

for (int pointsRowItr = 0; pointsRowItr < m; pointsRowItr++) {
}

int result = treeCoordinates(n, edges, points);

bufferedWriter.write(String.valueOf(result));
bufferedWriter.newLine();

bufferedWriter.close();
}
}

In Python3 :

#!/bin/python3

import sys

n,m = input().strip().split(' ')
n,m = [int(n),int(m)]
edges = []
for edges_i in range(n-1):
edges_t = [int(edges_temp) for edges_temp in input().strip().split(' ')]
edges.append(edges_t)
points = []
for points_i in range(m):
points_t = [int(points_temp) for points_temp in input().strip().split(' ')]
points.append(points_t)

class Node:
def __init__(self,i):
self.id = i
self.neighbors = set()

nodes = [Node(i) for i in range(n)]
for from_id,to_id in edges:

distances = [[10**5] * n for _ in range(n)]
for i in range(n):
distances[i][i] = 0

def set_distance(i, j, new_dist):
distances[i][j] = new_dist
distances[j][i] = new_dist

# find pair-wise distances with bfs <-- O(n^2)
for center_node in nodes:
cur_dist = 1
marked = set([center_node])
cur_level = set([center_node])
while cur_level:
next_level = set()
for node in cur_level:
for neighbor in node.neighbors:
if not neighbor in marked:
set_distance(center_node.id, neighbor.id, cur_dist)
cur_level = next_level
cur_dist += 1

# now go through the points
max_dist = 0
for p1 in range(m):
x1, y1 = points[p1]
x1 -= 1
y1 -= 1
for p2 in range(p1 + 1, m):
x2, y2 = points[p2]
x2 -= 1
y2 -= 1
new_dist = distances[x1][x2] + distances[y1][y2]
if new_dist > max_dist:
max_dist = new_dist

print(max_dist)

## Tree: Preorder Traversal

Complete the preorder function in the editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's preorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the preOrder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the tree's

## Tree: Postorder Traversal

Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the

## Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your \$inOrder* func

## Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary

## Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.

## Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F