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.

If node i has no left child then leftChild[i] will equal -1, similarly for the right child.

Note that the nodes have no values and that we only use the node numbers in this problem.

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

Constraints:

• 1 <= n <= 10^4

• leftChild.length == rightChild.length == n

• -1 <= leftChild[i], rightChild[i] <= n - 1

class Solution {
    public boolean validateBinaryTreeNodes(int n, int[] leftChild, int[] rightChild) {
        int[] parent = new int[n];
        for (int i = 0; i < n; i++)
            parent[i] = i;
        for (int i = 0; i < n; i++) {
            // Things taken care of: Cycles
            if (leftChild[i] != -1)
                if (union(parent, i, leftChild[i]))
                    return false;
            if (rightChild[i] != -1)
                if (union(parent, i, rightChild[i]))
                    return false;
        }
        // For taking care of multiple heads
        int count = 0;
        for (int i = 0; i < n; i++)
            if (parent[i] == i)
                count++;
        return count == 1;
    }

    public boolean union(int[] parent, int node1, int node2) {
        int p1 = find(parent, node1);
        int p2 = find(parent, node2);
        if (p1 == p2)
            return true;
        parent[p1] = p2;
        return false;
    }

    public int find(int[] parent, int node) {
        if (node != parent[node])
            parent[node] = find(parent, parent[node]);
        return parent[node];
    }
}