動機

把需要的spec定義出來就好

Problem

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 <= 104
  • leftChild.length == rightChild.length == n
  • -1 <= leftChild[i], rightChild[i] <= n - 1

Sol

  1. 只有一個root
  2. 沒有loop
  3. 所有點都要被走過
class Solution:
    def validateBinaryTreeNodes(self, n: int, leftChild: List[int], rightChild: List[int]) -> bool:
        root = set(range(n)) - (set(leftChild) | set(rightChild))
        
        if len(root) != 1:
            return False
        else:
            root = list(root)[0]
            seen = set()
            
            def dfs(i):
                if i == -1:
                    return True
                elif i in seen:
                    return False
                else:
                    seen.add(i)
                    return dfs(leftChild[i]) and dfs(rightChild[i])
            return dfs(root) and len(seen) == n