Find the smallest set of vertices from which all nodes in the graph are reachable. It’s guaranteed that a unique solution exists.

Notice that you can return the vertices in any order.

Example 1:

Input: n = 6, edges = [[0,1],[0,2],[2,5],[3,4],[4,2]]
Output: [0,3]
Explanation: It's not possible to reach all the nodes from a single vertex. From 0 we can reach [0,1,2,5]. From 3 we can reach [3,4,2,5]. So we output [0,3].

Example 2:

Input: n = 5, edges = [[0,1],[2,1],[3,1],[1,4],[2,4]]
Output: [0,2,3]
Explanation: Notice that vertices 0, 3 and 2 are not reachable from any other node, so we must include them. Also any of these vertices can reach nodes 1 and 4.

Constraints:

• 2 <= n <= 10^5
• 1 <= edges.length <= min(10^5, n * (n - 1) / 2)
• edges[i].length == 2
• 0 <= fromi, toi < n
• All pairs (fromi, toi) are distinct.

Solution: In degree

Nodes with 0 in degree will be the answer.
Time complexity: O(E+V)
Space complexity: O(V)

class Solution {
public:
  vector<int> findSmallestSetOfVertices(int n, vector<vector<int>>& edges) {
    vector<int> in(n);    
    for (const auto& e : edges) ++in[e[1]];
    vector<int> ans;
    for (int i = 0; i < n; ++i)
      if (in[i] == 0) ans.push_back(i);
    return ans;    
  }
};

The post 花花酱 LeetCode 1557. Minimum Number of Vertices to Reach All Nodes appeared first on Huahua's Tech Road.

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

Solution: Count in-degrees for each node

in degree must <= 1 and there must be exact one node that has 0 in-degree.

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

// Author: Huahua
class Solution {
public:
  bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
    vector<int> in(n);
    for (int l : leftChild) if (l >= 0) ++in[l];
    for (int r : rightChild) if (r >= 0) ++in[r];
    int zero = 0;
    for (int i = 0; i < n; ++i) {
      if (in[i] > 1) return false;
      if (in[i] == 0) ++zero;
    }
    // # of nodes without parent must be 1.
    return zero == 1;
  }
};

The post 花花酱 LeetCode 1361. Validate Binary Tree Nodes appeared first on Huahua's Tech Road.