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.
1var validateBinaryTreeNodes = function(n, leftChild, rightChild) {
2 const parent = Array.from({ length: n }, (_, index) => index);
3 const find = (x) => {
4 if (parent[x] !== x) {
5 parent[x] = find(parent[x]);
6 }
7 return parent[x];
8 };
9
10 const union = (x, y) => {
11 const rootX = find(x);
12 const rootY = find(y);
13 if (rootX === rootY) return false;
14 parent[rootY] = rootX;
15 return true;
16 };
17
18 const hasParent = new Array(n).fill(0);
19
20 for (let i = 0; i < n; ++i) {
21 if (leftChild[i] !== -1) {
22 if (hasParent[leftChild[i]] === 1 || !union(i, leftChild[i])) {
23 return false;
24 }
25 hasParent[leftChild[i]] = 1;
26 }
27 if (rightChild[i] !== -1) {
28 if (hasParent[rightChild[i]] === 1 || !union(i, rightChild[i])) {
29 return false;
30 }
31 hasParent[rightChild[i]] = 1;
32 }
33 }
34
35 return hasParent.filter(parent => parent === 0).length === 1;
36};
The JavaScript variation employs similar structures to monitor the tree integrity. Using find
and union
functions, this approach enforces the hierarchical conditions of nodes along with a snapshot of their root relations. In JavaScript, this procedural logic ensures the union-find checks can prevent improper cycles or multiple root scenarios.
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.
1public class Solution {
2 private void Dfs(int node, bool[] visited, int[][] children) {
3 if (visited[node]) return;
4 visited[node] = true;
5 if (children[0][node] != -1) Dfs(children[0][node], visited, children);
6 if (children[1][node] != -1) Dfs(children[1][node], visited, children);
7 }
8
9 public bool ValidateBinaryTreeNodes(int n, int[] leftChild, int[] rightChild) {
10 int[] inDegree = new int[n];
11 for (int i = 0; i < n; i++) {
12 if (leftChild[i] != -1) inDegree[leftChild[i]]++;
13 if (rightChild[i] != -1) inDegree[rightChild[i]]++;
14 }
15
16 int root = -1;
17 for (int i = 0; i < n; i++) {
18 if (inDegree[i] == 0) {
19 if (root == -1) {
20 root = i;
21 } else {
22 return false;
23 }
24 }
25 }
26
27 if (root == -1) return false;
28
29 bool[] visited = new bool[n];
30 Dfs(root, visited, new int[][] { leftChild, rightChild });
31
32 foreach (bool v in visited) {
33 if (!v) return false;
34 }
35
36 return true;
37 }
38}
The C# approach determines the root node based on the inDegree
, working under the trademark binary tree outline that each node may only have a single incoming edge (or reference) unless it is the root. Exploiting this relation, all nodes should receive connectivity verification through Dfs
, making sure the graph satisfies full extent reachability in step-results.