1448.count-good-nodes-in-binary-tree.py

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# @lc app=leetcode id=1448 lang=python3
#
# [1448] Count Good Nodes in Binary Tree
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# https://leetcode.com/problems/count-good-nodes-in-binary-tree/description/
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# algorithms
# Medium (73.30%)
# Likes:    6024
# Dislikes: 193
# Total Accepted:    660.2K
# Total Submissions: 900.7K
# Testcase Example:  '[3,1,4,3,null,1,5]'
#
# 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.
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# Return the number of good nodes in the binary tree.
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#
# Example 1:
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# 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.
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# Example 2:
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# Input: root = [3,3,null,4,2]
# Output: 3
# Explanation: Node 2 -> (3, 3, 2) is not good, because "3" is higher than it.
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# Example 3:
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# Input: root = [1]
# Output: 1
# Explanation: Root is considered as good.
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# Constraints:
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# 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].
#
#

# @lc code=start
# Definition for a binary tree node.

class TreeNode:
    def __init__(self, val=0, left=None, right=None, max_val=None):
        self.val = val
        self.left = left
        self.right = right
        self.max_val = max_val


class Solution:
    def goodNodes(self, root: TreeNode) -> int:
        root.max_val = root.val
        stack = [root]
        result = 1
        while stack:
            n = stack.pop()
            if n:
                # print(n.max_val)
                if n.left and n.left.val >= n.max_val:
                    result += 1
                    n.left.max_val = n.left.val
                elif n.left:
                    n.left.max_val = n.max_val
                if n.right and n.right.val >= n.max_val:
                    result += 1
                    n.right.max_val = n.right.val
                elif n.right:
                    n.right.max_val = n.max_val
                stack.append(n.left)
                stack.append(n.right)
                # print(n.val)
        return result


# @lc code=end