Introduction

In this blog post, we will explore how to solve the LeetCode problem "1448" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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,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. Example 2:
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].

Explanation

Here's the breakdown of the solution:

• DFS with Maximum Value Tracking: Perform a Depth-First Search (DFS) traversal of the binary tree. During the traversal, maintain the maximum value encountered along the path from the root to the current node.
• Good Node Check: At each node, compare its value with the maximum value seen so far. If the node's value is greater than or equal to the maximum value, it's a "good" node, and we increment the count.

• Update Maximum Value: Before exploring the left and right subtrees of a node, update the maximum value to be the larger of the current maximum and the node's value. Pass this updated maximum value to the recursive calls for the subtrees.

• Time Complexity: O(N), where N is the number of nodes in the tree.

• Space Complexity: O(H), where H is the height of the tree (worst case O(N) for skewed tree, average case O(log N) for balanced tree). This space is used by the recursion stack.

Code

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

def goodNodes(root: TreeNode) -> int:
    """
    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.
    """

    def dfs(node: TreeNode, max_val: int) -> int:
        """
        Performs a Depth-First Search to count good nodes.
        """
        if not node:
            return 0

        count = 0
        if node.val >= max_val:
            count = 1

        new_max_val = max(max_val, node.val)
        count += dfs(node.left, new_max_val)
        count += dfs(node.right, new_max_val)

        return count

    return dfs(root, root.val)