# Tree Detection - Facebook Top Interview Questions

### Problem Statement :

```You are given two lists of integers left and right, both of them the same length and representing a directed graph.

left[i] is the index of node i's left child and right[i] is the index of node i's right child.

A null child is represented by -1.

Return whether left and right represents a binary tree.

Constraints

n ≤ 100,000 where n is the length of left and right

Example 1

Input

left = [1, -1, 3, -1]

right = [2, -1, -1, -1]

Output

True

Example 2

Input

left = 

right = 

Output

False

Explanation

This is a circular node.```

### Solution :

```                        ```Solution in C++ :

int count_nodes(int root, vector<int>& left, vector<int>& right) {
if (root == -1) return 0;

return 1 + count_nodes(left[root], left, right) + count_nodes(right[root], left, right);
}

bool solve(vector<int>& left, vector<int>& right) {  // Time and Space: O(N)
// No cell can have indegree more than 1
int n = left.size();
vector<int> indeg(n, 0);

for (int i = 0; i < n; i++) {
if (left[i] != -1) {
if (++indeg[left[i]] > 1) return false;
}

if (right[i] != -1) {
if (++indeg[right[i]] > 1) return false;
}
}

// It has to have a root
int root = -1;
for (int i = 0; i < n; i++) {
if (indeg[i] == 0) {
root = i;
break;
}
}

if (root == -1) return false;

// It has to be connected
return count_nodes(root, left, right) == n;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
private Map<Integer, List<Integer>> graph = new HashMap();
private Set<Integer> visited = new HashSet();
private Set<Integer> visiting = new HashSet();
private boolean isCycleFound = false;
public boolean solve(int[] left, int[] right) {
int vertices = left.length;
int[] inDegreeMap = new int[left.length];
boolean isZeroIndegreeNodeFound = false;
int rootVertex = -1;
for (int i = 0; i < left.length; i++) {
if (left[i] != -1) {
inDegreeMap[left[i]]++;
}
if (right[i] != -1) {
inDegreeMap[right[i]]++;
}
}

for (int i = 0; i < inDegreeMap.length; ++i) {
if (inDegreeMap[i] > 1)
return false;
if (inDegreeMap[i] == 0 && isZeroIndegreeNodeFound)
return false;
if (inDegreeMap[i] == 0 && !isZeroIndegreeNodeFound) {
isZeroIndegreeNodeFound = true;
rootVertex = i;
}
}

if (rootVertex == -1)
return false;
dfs(rootVertex);
return (!isCycleFound && visited.size() == vertices);
}

private void dfs(int vertex) {
if (graph.containsKey(vertex)) {
for (int adjacentVertex : graph.get(vertex)) {
isCycleFound = true;
return;
}
}
}
visiting.remove(vertex);
}

private void addEdge(int u, int v) {
List<Integer> adjList = graph.getOrDefault(u, new ArrayList());
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, left, right):
N = len(left)
degs =  * N
for i in range(N):
if left[i] != -1:
degs[left[i]] += 1
if right[i] != -1:
degs[right[i]] += 1

q = deque(i for i in range(N) if degs[i] == 0)
if len(q) != 1:
return False

seen = set()
while q:
cur = q.popleft()
if cur in seen:
return False
if left[cur] != -1:
q.append(left[cur])
if right[cur] != -1:
q.append(right[cur])

return len(seen) == N```
```

## Square-Ten Tree

The square-ten tree decomposition of an array is defined as follows: The lowest () level of the square-ten tree consists of single array elements in their natural order. The level (starting from ) of the square-ten tree consists of subsequent array subsegments of length in their natural order. Thus, the level contains subsegments of length , the level contains subsegments of length , the

## Balanced Forest

Greg has a tree of nodes containing integer data. He wants to insert a node with some non-zero integer value somewhere into the tree. His goal is to be able to cut two edges and have the values of each of the three new trees sum to the same amount. This is called a balanced forest. Being frugal, the data value he inserts should be minimal. Determine the minimal amount that a new node can have to a

## Jenny's Subtrees

Jenny loves experimenting with trees. Her favorite tree has n nodes connected by n - 1 edges, and each edge is ` unit in length. She wants to cut a subtree (i.e., a connected part of the original tree) of radius r from this tree by performing the following two steps: 1. Choose a node, x , from the tree. 2. Cut a subtree consisting of all nodes which are not further than r units from node x .

## Tree Coordinates

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

## Array Pairs

Consider an array of n integers, A = [ a1, a2, . . . . an] . Find and print the total number of (i , j) pairs such that ai * aj <= max(ai, ai+1, . . . aj) where i < j. Input Format The first line contains an integer, n , denoting the number of elements in the array. The second line consists of n space-separated integers describing the respective values of a1, a2 , . . . an .

## Self Balancing Tree

An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. We define balance factor for each node as : balanceFactor = height(left subtree) - height(righ