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Formatted question description: https://leetcode.ca/all/1361.html
1361. Validate Binary Tree Nodes (Medium)
You have n binary tree nodes numbered from 0 to n - 1 where node i has two children leftChild[i] and rightChild[i], return true if and only if all the given nodes form exactly one valid binary tree.
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 <= 10^4leftChild.length == rightChild.length == n-1 <= leftChild[i], rightChild[i] <= n - 1
Related Topics:
Graph
Solution 1.
Firstly, we can check the in-dgree.
- Only one node (the root) has 0 in-degree. All other nodes have 1 in-dgree.
But this is not enough. For example:
[0,-1]
[-1,-1]
This graphy has one circle and one dangling node and satisfies the above requirement.
To rule this scenario out, we can add this requirement.
- If there are more than one node, the node with 0 in-degree must have out-degree.
// OJ: https://leetcode.com/problems/validate-binary-tree-nodes/
// Time: O(N)
// Space: O(N)
class Solution {
public:
bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
vector<int> indegree(n);
for (int i = 0; i < n; ++i) {
int L = leftChild[i], R = rightChild[i];
if (L != -1 && indegree[L]++) return false; // if the indegree is more than 1, return false
if (R != -1 && indegree[R]++) return false;
}
int root = -1;
for (int i = 0; i < n; ++i) {
if (indegree[i]) continue;
if (root != -1) return false; // if more than one node with 0 indegree, return false
root = i;
}
return root != -1 && (n == 1 || leftChild[root] != -1 || rightChild[root] != -1); // root must exist, and the root must have out-degree unless there is only one node
}
};
-
class Solution { public boolean validateBinaryTreeNodes(int n, int[] leftChild, int[] rightChild) { int[] parents = new int[n]; Arrays.fill(parents, -1); int edgesCount = 0; for (int i = 0; i < n; i++) { int curLeft = leftChild[i], curRight = rightChild[i]; if (curLeft >= 0) { edgesCount++; if (parents[curLeft] < 0) parents[curLeft] = i; else return false; } if (curRight >= 0) { edgesCount++; if (parents[curRight] < 0) parents[curRight] = i; else return false; } } if (edgesCount != n - 1) return false; int root = -1; for (int i = 0; i < n; i++) { if (parents[i] == -1) { if (root < 0) root = i; else return false; } } if (root < 0) return false; int visitCount = 0; Queue<Integer> queue = new LinkedList<Integer>(); queue.offer(root); while (!queue.isEmpty()) { int node = queue.poll(); visitCount++; int left = leftChild[node], right = rightChild[node]; if (left >= 0) queue.offer(left); if (right >= 0) queue.offer(right); } return visitCount == n; } } ############ class Solution { private int[] p; public boolean validateBinaryTreeNodes(int n, int[] leftChild, int[] rightChild) { p = new int[n]; for (int i = 0; i < n; ++i) { p[i] = i; } boolean[] vis = new boolean[n]; for (int i = 0, m = n; i < m; ++i) { for (int j : new int[] {leftChild[i], rightChild[i]}) { if (j != -1) { if (vis[j] || find(i) == find(j)) { return false; } p[find(i)] = find(j); vis[j] = true; --n; } } } return n == 1; } private int find(int x) { if (p[x] != x) { p[x] = find(p[x]); } return p[x]; } } -
// OJ: https://leetcode.com/problems/validate-binary-tree-nodes/ // Time: O(N) // Space: O(N) class Solution { public: bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) { vector<int> indegree(n); for (int i = 0; i < n; ++i) { int L = leftChild[i], R = rightChild[i]; if (L != -1 && indegree[L]++) return false; // if the indegree is more than 1, return false if (R != -1 && indegree[R]++) return false; } int root = -1; for (int i = 0; i < n; ++i) { if (indegree[i]) continue; if (root != -1) return false; // if more than one node with 0 indegree, return false root = i; } return root != -1 && (n == 1 || leftChild[root] != -1 || rightChild[root] != -1); // root must exist, and the root must have out-degree unless there is only one node } }; -
class Solution: def validateBinaryTreeNodes( self, n: int, leftChild: List[int], rightChild: List[int] ) -> bool: p = list(range(n)) vis = [False] * n def find(x): if p[x] != x: p[x] = find(p[x]) return p[x] for i in range(n): l, r = leftChild[i], rightChild[i] if l != -1: if vis[l] or find(i) == find(l): return False p[find(i)] = find(l) vis[l] = True n -= 1 if r != -1: if vis[r] or find(i) == find(r): return False p[find(i)] = find(r) vis[r] = True n -= 1 return n == 1 -
func validateBinaryTreeNodes(n int, leftChild []int, rightChild []int) bool { p := make([]int, n) for i := range p { p[i] = i } var find func(int) int find = func(x int) int { if p[x] != x { p[x] = find(p[x]) } return p[x] } vis := make([]bool, n) for i, a := range leftChild { for _, j := range []int{a, rightChild[i]} { if j != -1 { if vis[j] || find(i) == find(j) { return false } p[find(i)] = find(j) vis[j] = true n-- } } } return n == 1 }