1361. Validate Binary Tree Nodes (M)

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

题意

给定n个结点以及每个结点对应的左右子树信息，判断这些结点构成的图是否是一个有效的二叉树。

思路

若一个图满足以下三个要求，就是一个有效的二叉树：

1. 有且仅有一个结点的入度为0；
2. 所有结点的入度最大为1；
3. 所有结点的出度最大为2(题目本身就满足)。

代码实现

class Solution {
    public boolean validateBinaryTreeNodes(int n, int[] leftChild, int[] rightChild) {
        int[] in = new int[n];
        // 生成入度数组
        for (int i = 0; i < n; i++) {
            if (leftChild[i] != -1) {
                in[leftChild[i]]++;
            }
            if (rightChild[i] != -1) {
                in[rightChild[i]]++;
            }
        }
        int count = 0;
        for (int i = 0; i < n; i++) {
            // 判断最大入度数
            if (in[i] > 1) {
                return false;
            }
            // 统计入度为0的结点数
            if (in[i] == 0) {
                count++;
            }
        }
        if (count != 1) {
            return false;
        }
        return true;
    }
}