**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 = [0] right = [0] 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]]++;
addEdge(i, left[i]);
}
if (right[i] != -1) {
inDegreeMap[right[i]]++;
addEdge(i, 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) {
visiting.add(vertex);
if (graph.containsKey(vertex)) {
for (int adjacentVertex : graph.get(vertex)) {
if (visiting.contains(adjacentVertex)) {
isCycleFound = true;
return;
}
dfs(adjacentVertex);
}
}
visiting.remove(vertex);
visited.add(vertex);
}
private void addEdge(int u, int v) {
List<Integer> adjList = graph.getOrDefault(u, new ArrayList());
adjList.add(v);
graph.put(u, adjList);
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, left, right):
N = len(left)
degs = [0] * 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
seen.add(cur)
if left[cur] != -1:
q.append(left[cur])
if right[cur] != -1:
q.append(right[cur])
return len(seen) == N
```

## View More Similar Problems

## Balanced Brackets

A bracket is considered to be any one of the following characters: (, ), {, }, [, or ]. Two brackets are considered to be a matched pair if the an opening bracket (i.e., (, [, or {) occurs to the left of a closing bracket (i.e., ), ], or }) of the exact same type. There are three types of matched pairs of brackets: [], {}, and (). A matching pair of brackets is not balanced if the set of bra

View Solution →## Equal Stacks

ou have three stacks of cylinders where each cylinder has the same diameter, but they may vary in height. You can change the height of a stack by removing and discarding its topmost cylinder any number of times. Find the maximum possible height of the stacks such that all of the stacks are exactly the same height. This means you must remove zero or more cylinders from the top of zero or more of

View Solution →## Game of Two Stacks

Alexa has two stacks of non-negative integers, stack A = [a0, a1, . . . , an-1 ] and stack B = [b0, b1, . . . , b m-1] where index 0 denotes the top of the stack. Alexa challenges Nick to play the following game: In each move, Nick can remove one integer from the top of either stack A or stack B. Nick keeps a running sum of the integers he removes from the two stacks. Nick is disqualified f

View Solution →## Largest Rectangle

Skyline Real Estate Developers is planning to demolish a number of old, unoccupied buildings and construct a shopping mall in their place. Your task is to find the largest solid area in which the mall can be constructed. There are a number of buildings in a certain two-dimensional landscape. Each building has a height, given by . If you join adjacent buildings, they will form a solid rectangle

View Solution →## Simple Text Editor

In this challenge, you must implement a simple text editor. Initially, your editor contains an empty string, S. You must perform Q operations of the following 4 types: 1. append(W) - Append W string to the end of S. 2 . delete( k ) - Delete the last k characters of S. 3 .print( k ) - Print the kth character of S. 4 . undo( ) - Undo the last (not previously undone) operation of type 1 or 2,

View Solution →## Poisonous Plants

There are a number of plants in a garden. Each of the plants has been treated with some amount of pesticide. After each day, if any plant has more pesticide than the plant on its left, being weaker than the left one, it dies. You are given the initial values of the pesticide in each of the plants. Determine the number of days after which no plant dies, i.e. the time after which there is no plan

View Solution →