12.Validate Binary Tree Nodes

You have n binary tree nodes numbered from 0 to n - 1 where node i has two children leftChild[i] and rightChild[i], return true if and only if all the given nodes form exactly one valid binary tree.

Example 1:
Input: n = 4, leftChild = [1,-1,3,-1], rightChild = [2,-1,-1,-1]
Output: true

Example 2:
Input: n = 4, leftChild = [1,-1,3,-1], rightChild = [2,3,-1,-1]
Output: false


Example 3:
Input: n = 2, leftChild = [1,0], rightChild = [-1,-1]
Output: false

Example 4:
Input: n = 6, leftChild = [1,-1,-1,4,-1,-1], rightChild = [2,-1,-1,5,-1,-1]
Output: false

Solution : (Using BFS)

class Solution
{
public:
    bool validateBinaryTreeNodes(int n, vector<int> &leftChild, vector<int> &rightChild)
    {
        
        //finding parents
        vector<int> par(n);
        for (int i = 0; i < n; i++)
        {
            if (leftChild[i] != -1)
            {
                par[leftChild[i]]++;
            }

            if (rightChild[i] != -1)
            {
                par[rightChild[i]]++;
            }
        }
        
        // if A --> B exists then B --> A should not exist
        int node = -1;
        for (int i = 0; i < n; i++)
        {
            if (par[i] == 0)
            {
                node = i;
                break;
            }
        }
        if (node == -1)
        {
            return false;
        }
        
        //finding loop
        vector<bool> vs(n);
        queue<int> q;
        q.push(node);
        vs[node] = 1;

        while (!q.empty())
        {
            int x = q.front();
            q.pop();

            int lc = leftChild[x];
            if (lc != -1)
            {
                if (vs[lc] == 0)
                {
                    q.push(lc);
                    vs[lc] = 1;
                }
                else
                {
                    return false;
                }
            }

            int rc = rightChild[x];
            if (rc != -1)
            {
                if (vs[rc] == 0)
                {
                    q.push(rc);
                    vs[rc] = 1;
                }
                else
                {
                    return false;
                }
            }
        }

        //checking connectivity
        for (int i = 0; i < n; i++)
        {
            if (vs[i] == 0)
            {
                return false;
            }
        }

        return true;
    }
};