Today I have done leetcode's October LeetCoding Challenge with cpp.

October LeetCoding Challenge 17

Description

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

Constraints:

• n == leftChild.length == rightChild.length
• 1 <= n <= 104
• -1 <= leftChild[i], rightChild[i] <= n - 1

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
    vector<bool> root(n, true);
    for(int i = 0; i < n; ++i) {
      if(leftChild[i] != -1) root[leftChild[i]] = false;
      if(rightChild[i] != -1) root[rightChild[i]] = false;
    }
    queue<int> q;
    for(int i = 0; i < n; ++i) {
      if(root[i]) q.push(i);
    }
    if(q.size() != 1) return false;

    set<int> visit;
    while(q.size()) {
      auto current = q.front();
      q.pop();
      if(visit.count(current)) return false;
      visit.insert(current);
      if(leftChild[current] != -1) {
        q.push(leftChild[current]);
      }
      if(rightChild[current] != -1) {
        q.push(rightChild[current]);
      }
    }

    return visit.size() == n;
    
  }
};

// Accepted
// 44/44 cases passed (50 ms)
// Your runtime beats 46.81 % of cpp submissions
// Your memory usage beats 42.86 % of cpp submissions (37.5 MB)