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.
1class Solution {
2 public int find(int[] parent, int x) {
3 if (parent[x] != x) {
4 parent[x] = find(parent, parent[x]);
5 }
6 return parent[x];
7 }
8
9 public boolean unionFind(int[] parent, int x, int y) {
10 int rootX = find(parent, x);
11 int rootY = find(parent, y);
12
13 if (rootX == rootY) return false;
14 parent[rootY] = rootX;
15 return true;
16 }
17
18 public boolean validateBinaryTreeNodes(int n, int[] leftChild, int[] rightChild) {
19 int[] parent = new int[n];
20 int[] hasParent = new int[n];
21 for (int i = 0; i < n; ++i) {
22 parent[i] = i;
23 }
24
25 for (int i = 0; i < n; ++i) {
26 if (leftChild[i] != -1) {
27 if (hasParent[leftChild[i]] == 1 || !unionFind(parent, i, leftChild[i])) {
28 return false;
29 }
30 hasParent[leftChild[i]] = 1;
31 }
32 if (rightChild[i] != -1) {
33 if (hasParent[rightChild[i]] == 1 || !unionFind(parent, i, rightChild[i])) {
34 return false;
35 }
36 hasParent[rightChild[i]] = 1;
37 }
38 }
39
40 int rootCount = 0;
41 for (int i = 0; i < n; ++i) {
42 if (hasParent[i] == 0) {
43 rootCount++;
44 }
45 }
46
47 return rootCount == 1;
48 }
49}This Java solution, similar in logic to the C and C++ solutions, uses a union-find structure managed through unionFind and find methods. The hasParent array manages in-degree checks to ensure a single parent per node, avoiding tree invalidation by multiple roots or cycles. After scanning nodes, it confirms just one root using hasParent.
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.
1var validateBinaryTreeNodes = function(n, leftChild, rightChild) {
2 const inDegree = new Array(n).fill(0);
3
4 for (let i = 0; i < n; i++) {
5 if (leftChild[i] !== -1) inDegree[leftChild[i]]++;
6 if (rightChild[i] !== -1) inDegree[rightChild[i]]++;
7 }
8
9 let root = -1;
10 for (let i = 0; i < n; i++) {
11 if (inDegree[i] === 0) {
12 if (root === -1) {
13 root = i;
14 } else {
15 return false; // more than one root
16 }
17 }
18 }
19
20 if (root === -1) return false; // no root
21
22 const visited = new Array(n).fill(false);
23
24 const dfs = (node) => {
25 if (visited[node]) return;
26 visited[node] = true;
27
28 if (leftChild[node] !== -1) dfs(leftChild[node]);
29 if (rightChild[node] !== -1) dfs(rightChild[node]);
30 };
31
32 dfs(root);
33
34 return visited.every(v => v);
35};This JavaScript implementation similarly follows through verifying in-degree checks to locate the root and establishes a connection by guaranteeing complete DFS exploration from the root. This substantiates consistency in forming the tree in which connectivity is thoroughly enforced, extending checks across all nodes considered without overridden exceptions.