1361. Validate Binary Tree Nodes/树的判断

题目描述

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

解析

对于我这个小白来说，我觉得这是一道好题，让我学会了如何区分树与图。以下是题解中很简洁的一种做法。即利用树中不会有入度为2的结点，且如果这是一棵树而不是森林的话，入度为0的结点（根节点）只能有一个。

class Solution {
public:
    bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
        vector<int> inDegree(n,0);
        int count0=0,count2=0;
        for(int k:leftChild) if(k!=-1) inDegree[k]++;
        for(int k:rightChild) if(k!=-1) inDegree[k]++;
        for(int i=0;i<n;i++) {
            if(inDegree[i]==0) count0++;
            else if(inDegree[i]==2) count2++;
        }
        return count0==1&&count2==0;
    }
};