# Max Transform

### Problem Statement :

```Transforming data into some other data is typical of a programming job. This problem is about a particular kind of transformation which we'll call the max transform.

Let  be a zero-indexed array of integers. For , let  denote the subarray of  from index  to index , inclusive.

Let's define the max transform of  as the array obtained by the following procedure:

Let  be a list, initially empty.
For  from  to :
For  from  to :
Let .
Append  to the end of .
Return .
The returned array is defined as the max transform of . We denote it by .

Complete the function solve that takes an integer array  as input.

Given an array , find the sum of the elements of , i.e., the max transform of the max transform of . Since the answer may be very large, only find it modulo .

Input Format

The first line of input contains a single integer  denoting the length of .

The second line contains  space-separated integers  denoting the elements of .

Constraints

1  <=  n  <=  2x 10^5
1  <=  Ai  <= 10^6

Output Format

Print a single line containing a single integer denoting the answer.```

### Solution :

```                            ```Solution in C :

In   C++  :

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int ui;
typedef long double ld;
typedef pair<int, int> ii;
typedef pair<ii, ii> iii;
ll MOD = 1e9 + 7;
const ld E = 1e-9;
#define null NULL
#define ms(x) memset(x, 0, sizeof(x))
#ifndef LOCAL
#define endl "\n"
#endif
#ifndef LONG_LONG_MAX
#define LONG_LONG_MAX LLONG_MAX
#endif
#define sync ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define _write(x) freopen(x, "w", stdout)
#define files(x) _read(x ".in"); _write(x ".out")
#define filesdatsol(x) _read(x ".DAT"); _write(x ".SOL")
#define output _write("output.txt")
#define prev time_prev
#ifndef M_PI
#define M_PI acos(-1)
#endif
#define remove tipa_remove
#define next tipa_next
#define left tipa_left
#define right tipa_right
#define mod % MOD
#define y1 hello_world
unsigned char ccc;
bool _minus = false;
template<typename T>
inline T sqr(T t) {
return (t * t);
}
n = 0;
_minus = false;
while (true) {
ccc = getchar();
if (ccc == ' ' || ccc == '\n')
break;
if (ccc == '-') {
_minus = true;
continue;
}
n = n * 10 + ccc - '0';
}
if (_minus)
n *= -1;
}
n = 0;
_minus = false;
while (true) {
ccc = getchar();
if (ccc == ' ' || ccc == '\n') {
if (ccc == '\n')
return true;
break;
}
if (ccc == '-') {
_minus = true;
continue;
}
n = n * 10 + ccc - '0';
}
if (_minus)
n *= -1;
return false;
}
char wwww[19];
int kkkk;
inline void write(ll y) {
long long x = y;
kkkk = 0;
if (x < 0) {
putchar('-');
x *= -1;
}
if (!x)
wwww[++kkkk] = '0';
else
while (x) {
++kkkk;
wwww[kkkk] = char(x % 10 + '0');
x /= 10;
}
for (int i = kkkk; i >= 1; --i)
putchar(wwww[i]);
}
#ifdef LOCAL
//#define __DEBUG
#endif
#ifdef __DEBUG
#define dbg if(1)
#else
#define dbg if(0)
#endif

inline ll sum(ll n){
return (n * (n + 1)) / 2;
}

__int128 ans;

const int MAX = 4e5 + 10;

int ar[MAX];
ll tt[MAX];
ll ttt[MAX];
int n;

ll culc(int a){
return tt[a] % MOD;
}

int t[MAX << 2];

void build(int v, int l, int r){
if(l == r){
t[v] = l;
return;
}
int x = (l + r) >> 1;
build(v << 1, l, x);
build(v << 1 | 1, x + 1, r);
t[v] = (ar[t[v << 1]] > ar[t[v << 1 | 1]] ? t[v << 1] : t[v << 1 | 1]);
}

int get_max(int v, int tl, int tr, int l, int r){
if(l <= tl && tr <= r){
return t[v];
}
if(tr < l || r < tl){
return 0;
}
int x = (tl + tr) >> 1;
int a = get_max(v << 1, tl, x, l, r);
int b = get_max(v << 1 | 1, x + 1, tr, l, r);
return (ar[a] > ar[b] ? a : b);
}

ll culc(int a, int b){
ll ans = sum(a);
a--;
int r = a + b;
int l = a + b - 2 * min(a, b);
l = max(l, 0);
ans += ttt[r] - ttt[l];
int e = min(a, b);
a -= e;
b -= e;
ans += culc(a) + culc(b);
return ans;
}

int get_max(int l, int r){
return get_max(1, 1, n, l, r);
}

int get_max(int l1, int r1, int l2, int r2){
int a = get_max(l1, r1);
int b = get_max(l2, r2);
return (ar[a] > ar[b] ? a : b);
}

ll solve_1(int l, int r){
if(l > r)
return 0;
if(l == r){
ans += ar[l];
return 1;
}
int pos = get_max(l, r);
ll has = culc(r - l + 1);
has -= solve_1(l, pos - 1);
has -= solve_1(pos + 1, r);
has %= MOD;
ans += has * ar[pos];
return culc(r - l + 1);
}

ll solve_2(int r, int l){
if(r == 0){
return solve_1(l, n);
}
if(l == n + 1){
return solve_1(1, r);
}
int pos = get_max(1, r, l, n);
ll has = culc(r, n - l + 1);
if(pos <= r){
has -= solve_2(pos - 1, l);
has -= solve_1(pos + 1, r);
}else{
has -= solve_2(r, pos + 1);
has -= solve_1(l, pos - 1);
}
has %= MOD;
ans += has * ar[pos];
return culc(r, n - l + 1);
}

namespace solve_long {
vector<int> get(vector<int> v){
vector<int> t;
int n = (int) v.size();
for(int k = 0; k < n; k++){
for(int i = 0; i < n - k; i++){
int ans = 0;
for(int j = i; j <= i + k; j++){
ans = max(ans, v[j]);
}
t.push_back(ans);
}
}
return t;
}
int max_val = 0;
int get_max(vector<int> &v, int l, int r){
int pos = l;
for(int i = l + 1; i <= r && v[pos] != max_val; i++){
if(v[i] > v[pos]){
pos = i;
}
}
return pos;
}
ll sum(vector<int> &v, int l, int r){
if(l > r)
return 0;
int pos = get_max(v, l, r);
ll ans = (pos - l + 1) * 1LL * (r - pos + 1) * v[pos];
return ans + sum(v, l, pos - 1) + sum(v, pos + 1, r);
}

ll sum(vector<int> v){
max_val = 0;
for(int a : v){
max_val = max(max_val, a);
}
return sum(v, 0, (int) v.size() - 1);
}
__int128 solve(vector<int> v){
return sum((get(v))) % MOD;
}
}

ostream& operator<<(ostream &cout, __int128 a){
string s = "";
while(a){
s += char(a % 10 + '0');
a /= 10;
}
reverse(s.begin(), s.end());
cout << s;
return cout;
}

__int128 solve_ok(vector<int> v){
n = (int) v.size();
for(int i = 1; i <= n; i++){
ar[i] = v[i - 1];
}
build(1, 1, n);
ans = 0;
__int128 g = n;
g = (g * (g + 1)) / 2;
g = (g * (g + 1)) / 2;

int pos = get_max(1, n);
g -= solve_2(pos - 1, pos + 1);

ans += g * ar[pos];
return ans % MOD;
}

int main() {
sync;
srand((unsigned int) time(NULL));
cout.precision(10);
cout << fixed;
#ifdef LOCAL
input;
output;
ll start = (ll) clock();
#else

#endif

for(int i = 1; i < MAX; i++){
tt[i] = sum(i) + tt[i - 1];
}
for(int i = 1; i < MAX; i++){
ttt[i] = sum(i);
if(i > 1){
ttt[i] += ttt[i - 2];
}
}

int n;
cin >> n;

vector<int> v(n);
for(int i = 0; i < n; i++){
cin >> v[i];
}
cout << solve_ok(v) << endl;

#ifdef LOCAL
cout << "=====" << endl;
cout << (clock() - start) * 1. / CLOCKS_PER_SEC << endl;
cout << clock() << endl;
cout << "=====" << endl;
#endif

// assert(false);

}

In   Java :

import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;

public class MaxTransform {
InputStream is;
PrintWriter out;
String INPUT = "";
int mod = 1000000007;

void solve()
{
int n = ni();
int[] a = na(n);
if(n == 1){
out.println(a[0]);
return;
}
long all = (long)n*(n+1)/2%mod;
all = all*(all+1)/2%mod;
for(int i = 1;i <= n-1;i++){
all -= (long)(n-i+1+n-i)*(n-i+1+n-i+1)/2;
if(i < n-1)all += (long)(n-i)*(n-i+1)/2;
all %= mod;
}
int amax = 0;
for(int v : a)amax = Math.max(amax, v);
long ans = all*amax;
ans %= mod;
int[] b = new int[n];
for(int i = 0;i < n;i++)b[i] = -a[i];
SegmentTreeRMQPos st = new SegmentTreeRMQPos(b);
imos = new long[n+3];
dfs(0, n, a, st);

for(int i = 0;i <= n+2;i++){
imos[i] %= mod;
}
for(int i = 0;i <= n+1;i++){
imos[i+1] += imos[i];
imos[i+1] %= mod;
}
for(int i = 0;i <= n+1;i++){
imos[i+1] += imos[i];
imos[i+1] %= mod;
}
for(int i = 1;i <= n;i++){
ans += (long)i*imos[i];
ans %= mod;
}

int[] sufs = new int[n];
for(int i = 0;i < n;i++){
sufs[i] = a[n-1-i];
if(i > 0)sufs[i] = Math.max(sufs[i], sufs[i-1]);
}
int[] pres = new int[n];
for(int i = 0;i < n;i++){
pres[i] = a[i];
if(i > 0)pres[i] = Math.max(pres[i], pres[i-1]);
}
long[] cpres = new long[n+1];
for(int i = 0;i < n;i++){
cpres[i+1] = cpres[i] + pres[i];
}
long[] csufs = new long[n+1];
for(int i = 0;i < n;i++){
csufs[i+1] = csufs[i] + sufs[i];
}

long temp = 0;
for(int i = n-1;i >= 1;i--){
if(i < n-1)temp += maxsum(sufs[i-1], pres, i+1, n, cpres);
temp += maxsum(pres[i], sufs, i-1, n, csufs);
ans += temp;
ans %= mod;
}
ans %= mod;
if(ans < 0)ans += mod;
out.println(ans);
}

long maxsum(int v, int[] a, int l, int r, long[] cum)
{
int ind = Arrays.binarySearch(a, l, r, v);
if(ind < 0)ind = -ind-1;
long ret = cum[r] - cum[ind] + (long)(ind-l) * v;
ret %= mod;
return ret;
}

long[] imos;

void dfs(int l, int r, int[] a, SegmentTreeRMQPos st)
{
if(l >= r)return;
st.minx(l, r);
int arg = st.minpos;
imos[1] += a[arg];
imos[arg-l+2] -= a[arg];
imos[r-arg+1] -= a[arg];
imos[r-l+2] += a[arg];
dfs(l, arg, a, st);
dfs(arg+1, r, a, st);
}

public static class SegmentTreeRMQPos {
public int M, H, N;
public int[] st;
public int[] pos;

public SegmentTreeRMQPos(int n)
{
N = n;
M = Integer.highestOneBit(Math.max(N-1, 1))<<2;
H = M>>>1;
st = new int[M];
pos = new int[M];
for(int i = 0;i < N;i++)pos[H+i] = i;
Arrays.fill(st, 0, M, Integer.MAX_VALUE);
for(int i = H-1;i >= 1;i--)propagate(i);
}

public SegmentTreeRMQPos(int[] a)
{
N = a.length;
M = Integer.highestOneBit(Math.max(N-1, 1))<<2;
H = M>>>1;
st = new int[M];
pos = new int[M];
for(int i = 0;i < N;i++){
st[H+i] = a[i];
pos[H+i] = i;
}
Arrays.fill(st, H+N, M, Integer.MAX_VALUE);
for(int i = H-1;i >= 1;i--)propagate(i);
}

public void update(int pos, int x)
{
st[H+pos] = x;
for(int i = (H+pos)>>>1;i >= 1;i >>>= 1)propagate(i);
}

private void propagate(int i)
{
if(st[2*i] <= st[2*i+1]){
st[i] = st[2*i];
pos[i] = pos[2*i];
}else{
st[i] = st[2*i+1];
pos[i] = pos[2*i+1];
}
}

public int minpos;
public int minval;

public int minx(int l, int r){
minval = Integer.MAX_VALUE;
minpos = -1;
if(l >= r)return minval;
while(l != 0){
int f = l&-l;
if(l+f > r)break;
int v = st[(H+l)/f];
if(v < minval){
minval = v;
minpos = pos[(H+l)/f];
}
l += f;
}

while(l < r){
int f = r&-r;
int v = st[(H+r)/f-1];
if(v < minval){
minval = v;
minpos = pos[(H+r)/f-1];
}
r -= f;
}
return minval;
}

public int min(int l, int r){
minpos = -1;
minval = Integer.MAX_VALUE;
min(l, r, 0, H, 1);
return minval;
}

private void min(int l, int r, int cl, int cr, int cur)
{
if(l <= cl && cr <= r){
if(st[cur] < minval){
minval = st[cur];
minpos = pos[cur];
}
}else{
int mid = cl+cr>>>1;
if(cl < r && l < mid)min(l, r, cl, mid, 2*cur);
if(mid < r && l < cr)min(l, r, mid, cr, 2*cur+1);
}
}
}

void run() throws Exception
{
is = INPUT.isEmpty() ? System.in :
new ByteArrayInputStream(INPUT.getBytes());
out = new PrintWriter(System.out);

long s = System.currentTimeMillis();
solve();
out.flush();
if(!INPUT.isEmpty())tr(
System.currentTimeMillis()-s+"ms");
}

public static void main(String[] args)
throws Exception { new MaxTransform().run();
}

private byte[] inbuf = new byte[1024];
public int lenbuf = 0, ptrbuf = 0;

{
if(lenbuf == -1)throw new InputMismatchException();
if(ptrbuf >= lenbuf){
ptrbuf = 0;
try { lenbuf = is.read(inbuf); }
catch (IOException e) { throw new InputMismatchException(); }
if(lenbuf <= 0)return -1;
}
return inbuf[ptrbuf++];
}

private boolean isSpaceChar(int c)
{ return !(c >= 33 && c <= 126); }
private int skip() { int b;
while((b = readByte()) != -1 &&
isSpaceChar(b)); return b; }

private double nd()
{ return Double.parseDouble(ns()); }
private char nc()
{ return (char)skip(); }

private String ns()
{
int b = skip();
StringBuilder sb = new StringBuilder();
while(!(isSpaceChar(b))){
sb.appendCodePoint(b);
}
return sb.toString();
}

private char[] ns(int n)
{
char[] buf = new char[n];
int b = skip(), p = 0;
while(p < n && !(isSpaceChar(b))){
buf[p++] = (char)b;
}
return n == p ? buf : Arrays.copyOf(buf, p);
}

private char[][] nm(int n, int m)
{
char[][] map = new char[n][];
for(int i = 0;i < n;i++)map[i] = ns(m);
return map;
}

private int[] na(int n)
{
int[] a = new int[n];
for(int i = 0;i < n;i++)a[i] = ni();
return a;
}

private int ni()
{
int num = 0, b;
boolean minus = false;
while((b = readByte()) != -1 &&
!((b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
}

while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
}
}

private long nl()
{
long num = 0;
int b;
boolean minus = false;
while((b = readByte()) != -1 && !(
(b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
}

while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
}
}

private static void tr(Object... o)
{ System.out.println(Arrays.deepToString(o)); }
}

In   C  :

#pragma GCC optimize ("Ofast")
#pragma GCC target ("sse4")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
const int mod = 1000000007, _2 = 500000004;
int N, MX = 0, tp, a[200010],
i_1[200010], st[200010], mxl[200010],
mxr[200010], sxl[200010], sxr[200010];
long long M, CNT, ANS = 0;
void calc(int w, int x, int y)
{
if( x < y )
{
int temp = x;
x = y;
y = temp;
}
int k;
if( x == y )
{
k = ( ( (long long)( x + y ) * i_1[y] % mod -
(long long)x * x % mod ) % mod + mod ) % mod;
}
else
{
k = ( ( (long long)y * ( i_1[x-1] - i_1[y] ) %
mod + (long long)( x + y ) * i_1[y] %
mod ) % mod + mod ) % mod;
}
ANS = ( ANS + (long long)w * k ) % mod;
CNT -= k;
if( CNT < 0 )
{
CNT += mod;
}
}
void calcl(int w, int x, int y)
{
if( x == 1 || y == 0 )
{
return;
}
int k;
if( y < x )
{
k = i_1[y];
}
else
{
k = ( i_1[x-1] + (long long)(
y - x + 1 ) * ( x - 1 ) ) % mod;
}
ANS = ( ANS + (long long)w * k ) % mod;
CNT -= k;
if( CNT < 0 )
{
CNT += mod;
}
}
void calcr(int w, int x, int y)
{
if( x == 0 || y == 1 )
{
return;
}
int k;
if( y + 1 <= x )
{
k = i_1[y-1];
}
else
{
k = ( i_1[x] + (long long)( y - x - 1 ) * x ) % mod;
}
ANS = ( ANS + (long long)w * k ) % mod;
CNT -= k;
if( CNT < 0 )
{
CNT += mod;
}
}
int main()
{
int p;
scanf("%d", &N);
for( int i = 1 ; i <= N ; i++ )
{
scanf("%d", &a[i]);
MX = MX > a[i] ? MX : a[i];
}
M = ( (long long)N *
( N + 1 ) >> 1 ) % mod;
M = (long long)M * ( M + 1 ) % mod * _2 % mod;
CNT = M;
for( int i = 1 ; i <= N ; i++ )
{
i_1[i] = ( i_1[i-1] + i ) % mod;
}
for( int i = 1 ; i <= N ; i++ )
{
sxl[i] = sxl[i-1] > a[i] ? sxl[i-1] : a[i];
}
for( int i = N ; i ; i-- )

{
sxr[i] = sxr[i+1] > a[i] ? sxr[i+1] : a[i];
}
tp = 0;
for( int i = 1 ; i <= N ; i++ )
{
while( tp > 0 && a[st[tp]] <= a[i] )
{
tp--;
}
if(tp)
{
mxl[i] = st[tp] + 1;
}
else
{
mxl[i] = 1;
}
st[++tp] = i;
}
tp = 0;
for( int i = N ; i ; i-- )
{
while( tp > 0 && a[st[tp]] < a[i] )
{
tp--;
}
if(tp)
{
mxr[i] = st[tp] - 1;
}
else
{
mxr[i] = N;
}
st[++tp] = i;
}
for( int i = 1 ; i <= N ; i++ )
{
calc(a[i], i-mxl[i]+1, mxr[i]-i+1);
}
p = N;
for( int i = 1 ; i <= N ; i++ )
{
int g = sxl[i];
while( p > i && sxr[p] < g )
{
p--;
}
while( p < i )
{
p++;
}
calcl(g, i, N-p);
}
p = 1;
for( int i = N ; i ; i-- )
{
int g = sxr[i];
while( p < i && sxl[p] <= g )
{
p++;
}
while( p > i )
{
p--;
}
calcr(g, N-i+1, p-1);
}
CNT = ( CNT % mod + mod ) % mod;
ANS = ( ANS + (long long)CNT * MX ) % mod;
printf("%lld", ANS);
return 0;
}

In   Python 3 :

#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the solve function below.

import math
import os
import random
import re
import sys
sys.setrecursionlimit(9999999)
from decimal import Decimal
def t1(n):
return Decimal(n * (n + 1) / 2)

def t2(n):
return Decimal(n * (n + 1) * (n + 2) / 6)

def u2(n):
return Decimal(n * (n + 2) * (2 * n + 5) / 24)

def countzip(a, b):
return u2(a + b) - u2(abs(a - b)) + t2(abs(a - b))

def countends(x, n, ex):
return countzip(n, ex) - countzip(x, ex) - countzip(n - 1 - x, 0)

def countsplit(x, n):
return t1(t1(n)) - t1(x) - countzip(n - x - 1, x - 1)

K = 20
lg = [0] * (1 << K)
for i in range(K):
lg[1 << i] = i
for i in range(1, 1 << K):
lg[i] = max(lg[i], lg[i - 1])

def make_rangemax(A):
n = len(A)
assert 1 << K > n

key = lambda x: A[x]
mxk = []
mxk.append(range(n))
for k in range(K - 1):
mxk.append(list(mxk[-1]))
for i in range(n - (1 << k)):
mxk[k + 1][i] = max(
mxk[k][i], mxk[k][i + (1 << k)],
key=key)

def rangemax(i, j):
k = lg[j - i]
return max(mxk[k][i], mxk[k][j - (1 << k)], key=key)

return rangemax

def brutesolo(A):
rangemax = make_rangemax(A)
stack = [(0, len(A))]
ans = 0
while stack:
i, j = stack.pop()
if i != j:
x = rangemax(i, j)
stack.append((i, x))
stack.append((x + 1, j))
ans += A[x] * (x - i + 1) * (j - x)
return ans

def make_brute(A):
rangemax = make_rangemax(A)

def brute(i, j):
stack = [(i, j)]
ans = 0
while stack:
i, j = stack.pop()
if i != j:
x = rangemax(i, j)
stack.append((i, x))
stack.append((x + 1, j))
ans += A[x] * countends(x - i, j - i, 0)
return ans

return brute, rangemax

def ends(A, B):
brutea, rangemaxa = make_brute(A)
bruteb, rangemaxb = make_brute(B)

stack = [(len(A), len(B))]
ans = 0
while stack:
i, j = stack.pop()
if i == 0:
ans += bruteb(0, j)
elif j == 0:
ans += brutea(0, i)
else:
x = rangemaxa(0, i)
y = rangemaxb(0, j)
if A[x] < B[y]:
ans += bruteb(y + 1, j)
ans += B[y] * countends(y, j, i)
stack.append((i, y))
else:
ans += brutea(x + 1, i)
ans += A[x] * countends(x, i, j)
stack.append((x, j))

return ans

def maxpairs(a):
return [max(x, y) for x, y in zip(a, a[1:])]

def solve(A):
n = len(A)
x = max(range(n), key=lambda x: A[x])
return (int((brutesolo(A[:x]) +
ends(A[x + 1:][::-1], maxpairs(A[:x])) +
A[x] * countsplit(x, n))%(10**9+7)))

if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')

n = int(input())

A = list(map(int, input().rstrip().split()))

result = solve(A)

fptr.write(str(result) + '\n')

fptr.close()```
```

## Binary Search Tree : Insertion

You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <

## Tree: Huffman Decoding

Huffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only t

## Binary Search Tree : Lowest Common Ancestor

You are given pointer to the root of the binary search tree and two values v1 and v2. You need to return the lowest common ancestor (LCA) of v1 and v2 in the binary search tree. In the diagram above, the lowest common ancestor of the nodes 4 and 6 is the node 3. Node 3 is the lowest node which has nodes and as descendants. Function Description Complete the function lca in the editor b

## Swap Nodes [Algo]

A binary tree is a tree which is characterized by one of the following properties: It can be empty (null). It contains a root node only. It contains a root node with a left subtree, a right subtree, or both. These subtrees are also binary trees. In-order traversal is performed as Traverse the left subtree. Visit root. Traverse the right subtree. For this in-order traversal, start from

## Kitty's Calculations on a Tree

Kitty has a tree, T , consisting of n nodes where each node is uniquely labeled from 1 to n . Her friend Alex gave her q sets, where each set contains k distinct nodes. Kitty needs to calculate the following expression on each set: where: { u ,v } denotes an unordered pair of nodes belonging to the set. dist(u , v) denotes the number of edges on the unique (shortest) path between nodes a

## Is This a Binary Search Tree?

For the purposes of this challenge, we define a binary tree to be a binary search tree with the following ordering requirements: The data value of every node in a node's left subtree is less than the data value of that node. The data value of every node in a node's right subtree is greater than the data value of that node. Given the root node of a binary tree, can you determine if it's also a