leetcode-1448 - Count Good Nodes in Binary Tree

動機

就dfs傳範圍

Problem

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,Leetcode,1,5]Output: 4Explanation: 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 pathNode 3 -> (3,1,3) is the maximum value in the path.

Example 2:

Input: root = [3,3,null,4,2]Output: 3Explanation: Node 2 -> (3, 3, 2) is not good, because "3" is higher than it.

Example 3:

Input: root = [1]Output: 1Explanation: 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].

Sol

class Solution:
    def goodNodes(self, root: TreeNode,val=float('-inf')) -> int:
        if not root:
            return 0
        elif root.val >= val:
            return 1+self.goodNodes(root.left,root.val)+self.goodNodes(root.right,root.val)
        else:
            return self.goodNodes(root.left,val)+self.goodNodes(root.right,val)