1361. Validate Binary Tree Nodes

Problem

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

  Constraints:

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

• 1 <= n <= 10^4

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

Solution (Java)

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<Integer> 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;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).