1448. Count Good Nodes in Binary Tree

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].

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# https://leetcode.com/problems/count-good-nodes-in-binary-tree/

'''
1. 아이디어 :
    1) dfs로 탐색하면서, 노드가 없으면 0, 노드의 값이 value보다 크거나 같으면 ans+=1. maxvalue를 현재 노드와 비교하여 갱신하고,
    자식 노드들도 dfs로 실행한다.
2. 시간복잡도 :
    1) O(n)
3. 자료구조 :
    1) 이진트리
'''


# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def goodNodes(self, root: TreeNode) -> int:
        ans = 0

        def dfs(node, value):
            nonlocal ans
            if not node:
                return 0
            if node.val >= value:
                ans+=1
            maxval = max(value,node.val)
            dfs(node.left,maxval)
            dfs(node.right,maxval)

        dfs(root,-float('inf'))
        return ans