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 :



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                        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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