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.
1class Solution:
2 def dfs(self, node, visited, children):
3 if visited[node]: return
4 visited[node] = True
5 if children[0][node] != -1:
6 self.dfs(children[0][node], visited, children)
7 if children[1][node] != -1:
8 self.dfs(children[1][node], visited, children)
9
10 def validateBinaryTreeNodes(self, n, leftChild, rightChild):
11 # Calculate in-degrees
12 inDegree = [0] * n
13 for i in range(n):
14 if leftChild[i] != -1:
15 inDegree[leftChild[i]] += 1
16 if rightChild[i] != -1:
17 inDegree[rightChild[i]] += 1
18
19 root = -1
20 for i in range(n):
21 if inDegree[i] == 0:
22 if root == -1:
23 root = i
24 else:
25 return False
26
27 if root == -1:
28 return False
29
30 visited = [False] * n
31 self.dfs(root, visited, [leftChild, rightChild])
32
33 return all(visited)
The Python strategem frames nodes as tuples within a list structure, exploiting recognized semantics over positions within binary arrays (left and right children). Node in-degrees determine the hierarchy without multi-parents obstruction, generating a singular possible valid root. DFS confirms full graph traversability from the root node, ensuring a connected nature without interruptions.