【leetcode】1361. Validate Binary Tree Nodes

题目如下：

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

解题思路：根据leftChild和rightChild的值模拟树的遍历，并记录遍历过的节点，最后判断节点数是否等于n。

代码如下：

class Solution(object):
    def validateBinaryTreeNodes(self, n, leftChild, rightChild):
        """
        :type n: int
        :type leftChild: List[int]
        :type rightChild: List[int]
        :rtype: bool
        """
        dic = {}
        
        queue = [0]
        while len(queue) > 0:
            node = queue.pop(0)
            if node in dic:
                return False
            dic[node] = 1
            if leftChild[node] != -1:
                queue.append(leftChild[node] )
            if rightChild[node] != -1:
                queue.append(rightChild[node] )
        return len(dic) == n