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]
.