leetcode 1448. 统计二叉树中好节点的数目

发表于 分类于 阅读次数：

遍历二叉树即可

时间复杂度：O(n)
空间复杂度：O(1)

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# 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 __init__(self) -> None:
        self.good_node_cnt = 0

    def goodNodes(self, root: TreeNode) -> int:
        self.good_node_cnt = 0
        self.solve_good_node_cnt(root, float('-INF'))
        return self.good_node_cnt
    
    def solve_good_node_cnt(self, root, max_val):
        if not root:
            return

        if root.val >= max_val:
            self.good_node_cnt += 1
            max_val = root.val
            
        self.solve_good_node_cnt(root.left, max_val)
        self.solve_good_node_cnt(root.right, max_val)