# 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]]++;
}
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 = [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
if left[cur] != -1:
q.append(left[cur])
if right[cur] != -1:
q.append(right[cur])

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

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