Count Good Nodes in Binary Tree

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- Tag: #binarytree, #dfs

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.

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.

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].

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

- Solution:

## https://code.dennyzhang.com/count-good-nodes-in-binary-tree ## Basic Ideas: pre-order ## Complexity: Time O(n), Space O(1) # 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: res = 0 def dfs(root, maxVal): nonlocal res if not root: return if root.val >= maxVal: res += 1 maxVal = root.val dfs(root.left, maxVal) dfs(root.right, maxVal) dfs(root, -sys.maxsize) return res