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 <= 10^4
leftChild.length == rightChild.length == n
-1 <= leftChild[i], rightChild[i] <= n - 1
Solution: Count in-degrees for each node
in degree must <= 1 and there must be exact one node that has 0 in-degree.
Time complexity: O(n)
Space complexity: O(n)
C++
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// Author: Huahua class Solution { public: bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) { vector<int> in(n); for (int l : leftChild) if (l >= 0) ++in[l]; for (int r : rightChild) if (r >= 0) ++in[r]; int zero = 0; for (int i = 0; i < n; ++i) { if (in[i] > 1) return false; if (in[i] == 0) ++zero; } // # of nodes without parent must be 1. return zero == 1; } }; |