LeetCode contest 177 5170. 验证二叉树
Table of Contents
中文版：
二叉树上有 n 个节点，按从 0 到 n - 1 编号，其中节点 i 的两个子节点分别是 leftChild[i] 和 rightChild[i]。
只有 所有 节点能够形成且 只 形成 一颗 有效的二叉树时，返回 true；否则返回 false。
如果节点 i 没有左子节点，那么 leftChild[i] 就等于 -1。右子节点也符合该规则。
注意：节点没有值，本问题中仅仅使用节点编号。
示例 1：
输入：n = 4, leftChild = [1,-1,3,-1], rightChild = [2,-1,-1,-1]
输出：true
示例 2：
输入：n = 4, leftChild = [1,-1,3,-1], rightChild = [2,3,-1,-1]
输出：false
示例 3：
输入：n = 2, leftChild = [1,0], rightChild = [-1,-1]
输出：false
示例 4：
输入：n = 6, leftChild = [1,-1,-1,4,-1,-1], rightChild = [2,-1,-1,5,-1,-1]
输出：false
提示：
1 <= n <= 10^4
leftChild.length == rightChild.length == n
-1 <= leftChild[i], rightChild[i] <= n - 1
英文版：
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
My answer:
# 算法：只有根节点没有父节点，其余节点都只有一个父节点。 class Solution: def validateBinaryTreeNodes(self, n: int, leftChild: List[int], rightChild: List[int]) -> bool: parent_list = [0] * n for i in range(len(leftChild)): if leftChild[i] != -1: parent_list[leftChild[i]] += 1 if rightChild[i] != -1: parent_list[rightChild[i]] += 1 print(parent_list) num_zero = 0 num_one = 0 num_other = 0 for x in parent_list: if x == 0: num_zero += 1 elif x == 1: num_one += 1 else: num_other += 1 return num_zero == 1 and num_one == n-1 and num_other == 0
代码优化后：
# 算法：只有根节点没有父节点，其余节点都只有一个父节点。 class Solution: def validateBinaryTreeNodes(self, n: int, leftChild: List[int], rightChild: List[int]) -> bool: parent_list = [0] * n for i in range(n): if leftChild[i] != -1: parent_list[leftChild[i]] += 1 if rightChild[i] != -1: parent_list[rightChild[i]] += 1 # print(parent_list) num_zero = 0 num_one = 0 # num_other = 0 for x in parent_list: if x == 0: num_zero += 1 elif x == 1: num_one += 1 # else: # num_other += 1 # return num_zero == 1 and num_one == n-1 and num_other == 0 return num_zero == 1 and num_one == n-1
解题报告：
1、如何判断二叉树
只有一个根节点，只有根节点没有父节点，其余节点只有一个父节点。
2、parent_list 代表 n 个节点是否有父节点，下标 index 代表 n 个节点，parent_list[index] 代表 index 这个节点有几个父节点。
