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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1301_1400.s1361_validate_binary_tree_nodes;

// #Medium #Depth_First_Search #Breadth_First_Search #Tree #Binary_Tree #Graph #Union_Find
// #2022_03_21_Time_8_ms_(69.78%)_Space_55.4_MB_(51.49%)

import java.util.ArrayDeque;
import java.util.Deque;

/**
 * 1361 - Validate Binary Tree Nodes\.
 *
 * Medium
 *
 * 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:**
 *
 * ![](https://assets.leetcode.com/uploads/2019/08/23/1503_ex1.png)
 *
 * **Input:** n = 4, leftChild = [1,-1,3,-1], rightChild = [2,-1,-1,-1]
 *
 * **Output:** true
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2019/08/23/1503_ex2.png)
 *
 * **Input:** n = 4, leftChild = [1,-1,3,-1], rightChild = [2,3,-1,-1]
 *
 * **Output:** false
 *
 * **Example 3:**
 *
 * ![](https://assets.leetcode.com/uploads/2019/08/23/1503_ex3.png)
 *
 * **Input:** n = 2, leftChild = [1,0], rightChild = [-1,-1]
 *
 * **Output:** false
 *
 * **Constraints:**
 *
 * *   `n == leftChild.length == rightChild.length`
 * *   1 <= n <= 10⁴
 * *   `-1 <= leftChild[i], rightChild[i] <= n - 1`
**/
public class Solution {
    public boolean validateBinaryTreeNodes(int n, int[] leftChild, int[] rightChild) {
        int[] inDeg = new int[n];
        for (int i = 0; i < n; i++) {
            if (leftChild[i] >= 0) {
                inDeg[leftChild[i]] += 1;
            }
            if (rightChild[i] >= 0) {
                inDeg[rightChild[i]] += 1;
            }
        }
        Deque queue = new ArrayDeque<>();
        for (int i = 0; i < n; i++) {
            if (inDeg[i] == 0) {
                if (queue.isEmpty()) {
                    queue.offer(i);
                } else {
                    // Violate rule 1.
                    return false;
                }
            }
            if (inDeg[i] > 1) {
                // Violate rule 2.
                return false;
            }
        }
        int tpLen = 0;
        while (!queue.isEmpty()) {
            int curNode = queue.poll();
            tpLen++;
            int left = leftChild[curNode];
            int right = rightChild[curNode];
            if (left > -1 && --inDeg[left] == 0) {
                queue.offer(left);
            }
            if (right > -1 && --inDeg[right] == 0) {
                queue.offer(right);
            }
        }
        return tpLen == n;
    }
}