Computer science note/LeetCode

Count Good Nodes in Binary Tree

problem description and solution for the problem "Count Good Nodes in Binary Tree" on LeetCode.

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Given a binary tree root, a node X in the tree is named good if in the path from root to X there are no nodes with a value greater than X.
给定一个二叉树,如果从根到X的路径中没有值大于X的节点,则树 root 中的节点X被命名为好。

Return the number of good nodes in the binary tree.
返回二叉树中良好节点的数量。

Example 1: 示例 1:

Input: root = [3,1,4,3,null,1,5] Output: 4 Explanation: Nodes in blue are good. Root Node (3) is always a good node. Node 4 -> (3,4) is the maximum value in the path starting from the root. Node 5 -> (3,4,5) is the maximum value in the path Node 3 -> (3,1,3) is the maximum value in the path.

Example 2: 示例 2:

Input: root = [3,3,null,4,2] Output: 3 Explanation: Node 2 -> (3, 3, 2) is not good, because "3" is higher than it.

Example 3: 例3:

Input: root = [1] Output: 1 Explanation: Root is considered as good.

Constraints: 约束:

  • The number of nodes in the binary tree is in the range [1, 10^5].
    二叉树中的节点数在 [1, 10^5] 范围内。
  • Each node's value is between [-10^4, 10^4].
    每个节点的值介于 [-10^4, 10^4] .