LeetCode 1448 Count Good Nodes in Binary Tree

LeetCode

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.

Return the number of good nodes in the binary tree.

Example 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:

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:

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

Constraints:

The number of nodes in the binary tree is in the range [1, 10^5]. Each node’s value is between [-10^4, 10^4].

题目大意：

一个节点是good表示该节点从root到自己的路径上，所有节点都小于等于自己。求二叉树的good节点个数

解题思路：

统计左右子树的good节点个数，最重要是引入类似于min, max验证BST，引入path_max来记录路径上的最大值，只要该节点值大于path_max就是good节点。DFS返回good节点个数

解题步骤：

N/A

注意事项：

1. 引入path_max来记录路径上的最大值

Python代码：

def goodNodes(self, root: TreeNode) -> int:
	return self.dfs(root, float('-inf'))

def dfs(self, root, path_max):
	if not root:
		return 0
	left = self.dfs(root.left, max(root.val, path_max))
	right = self.dfs(root.right, max(root.val, path_max))
	res = left + right
	if path_max <= root.val:
		res += 1
	return res

算法分析：

时间复杂度为O(n)，空间复杂度O(1)