To solve this problem using the Disjoint Set Union (DSU) approach, we aim to union nodes based on their parent-child relationships. A valid tree should follow these rules:
Time Complexity: O(n), where n is the number of nodes, due to each union and find operation being nearly constant with path compression.
Space Complexity: O(n) for the parent array.
1#include <vector>
2#include <unordered_set>
3
4class DSU {
5 std::vector<int> parent;
6public:
7 DSU(int n) {
8 parent.resize(n);
9 for (int i = 0; i < n; ++i) parent[i] = i;
10 }
11
12 int find(int x) {
13 if (parent[x] != x)
14 parent[x] = find(parent[x]);
15 return parent[x];
16 }
17
18 bool unionSets(int x, int y) {
19 int rootX = find(x), rootY = find(y);
20 if (rootX == rootY) return false;
21 parent[rootY] = rootX;
22 return true;
23 }
24};
25
26bool validateBinaryTreeNodes(int n, std::vector<int>& leftChild, std::vector<int>& rightChild) {
27 DSU dsu(n);
28 std::vector<int> inDegree(n, 0);
29
30 for (int i = 0; i < n; ++i) {
31 if (leftChild[i] != -1) {
32 if (++inDegree[leftChild[i]] > 1 || !dsu.unionSets(i, leftChild[i]))
33 return false;
34 }
35 if (rightChild[i] != -1) {
36 if (++inDegree[rightChild[i]] > 1 || !dsu.unionSets(i, rightChild[i]))
37 return false;
38 }
39 }
40
41 int rootCount = 0;
42 for (int i = 0; i < n; ++i) {
43 if (dsu.find(i) == i) ++rootCount;
44 }
45
46 return rootCount == 1;
47}This C++ solution is a close variant of the C example. It uses a class DSU structuring the union-find operations. The solution primarily checks for:
Vector inDegree tracks in-degrees, and the DSU tracks roots. After processing all nodes, the algorithm checks the single connected component from the root count.
This approach involves calculating the in-degree of each node and checking connectivity via a DFS. The key aspects of a tree like single-root presence and cycle-checking can be managed by:
Time Complexity: O(n), since each node and its immediate edges are evaluated once in each step, including in-drives calculations and DFS.
Space Complexity: O(n) for holding visited tracking and in-degree counts.
1#include <stdbool.h>
2#include <string.h>
3
4void dfs(int node, int* visited, int** children, int n) {
5 if (visited[node]) return;
6 visited[node] = 1;
7 if (children[0][node] != -1) dfs(children[0][node], visited, children, n);
8 if (children[1][node] != -1) dfs(children[1][node], visited, children, n);
9}
10
11bool validateBinaryTreeNodes(int n, int* leftChild, int leftChildSize, int* rightChild, int rightChildSize) {
12 int inDegree[n];
13 memset(inDegree, 0, sizeof(inDegree));
14
15 for (int i = 0; i < n; ++i) {
16 if (leftChild[i] != -1) ++inDegree[leftChild[i]];
17 if (rightChild[i] != -1) ++inDegree[rightChild[i]];
18 }
19
20 int root = -1;
21 for (int i = 0; i < n; ++i) {
22 if (inDegree[i] == 0) {
23 if (root == -1) {
24 root = i;
25 } else {
26 return false; // more than one root
27 }
28 }
29 }
30
31 if (root == -1) return false; // no root
32
33 int visited[n];
34 memset(visited, 0, sizeof(visited));
35 dfs(root, visited, (int*[]){leftChild, rightChild}, n);
36
37 for (int i = 0; i < n; ++i) {
38 if (!visited[i]) return false; // not fully connected
39 }
40
41 return true;
42}In this C solution, the algorithm starts by calculating the in-degree for each node, which should equal zero for the root node. A DFS checks connectivity from a root node, ensuring all nodes are part of one connected graph and are reachable. Each node is visited once, ensuring an entire traversal defines the single connected component rule of trees.