题目
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.
解题思路
DFS，每次将路径上的最大值传给下一层，并把左右子树的结果以及当前节点是否符合条件求出总和，返回给上一层
代码
class Solution:
def goodNodes(self, root: TreeNode) -> int:
def dfs(root, max_val):
res = int(root.val >= max_val)
if root.left:
res += dfs(root.left, max(max_val, root.val))
if root.right:
res += dfs(root.right, max(max_val, root.val))
return res
return dfs(root, -10000)
