**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
```

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