Introduction

In this blog post, we will explore how to solve the LeetCode problem "1609" 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

A binary tree is named Even-Odd if it meets the following conditions: The root of the binary tree is at level index 0, its children are at level index 1, their children are at level index 2, etc. For every even-indexed level, all nodes at the level have odd integer values in strictly increasing order (from left to right). For every odd-indexed level, all nodes at the level have even integer values in strictly decreasing order (from left to right). Given the root of a binary tree, return true if the binary tree is Even-Odd, otherwise return false. Example 1: Input: root = [1,10,4,3,null,7,9,12,8,6,null,null,2] Output: true Explanation: The node values on each level are: Level 0: [1] Level 1: [10,4] Level 2: [3,7,9] Level 3: [12,8,6,2] Since levels 0 and 2 are all odd and increasing and levels 1 and 3 are all even and decreasing, the tree is Even-Odd. Example 2: Input: root = [5,4,2,3,3,7] Output: false Explanation: The node values on each level are: Level 0: [5] Level 1: [4,2] Level 2: [3,3,7] Node values in level 2 must be in strictly increasing order, so the tree is not Even-Odd. Example 3: Input: root = [5,9,1,3,5,7] Output: false Explanation: Node values in the level 1 should be even integers. Constraints: The number of nodes in the tree is in the range [1, 105]. 1 <= Node.val <= 106

Explanation

Here's the solution to determine if a binary tree is Even-Odd:

• Level-Order Traversal: Use a level-order (breadth-first) traversal to visit nodes level by level.
• Even/Odd Check: For each level, check if the node values satisfy the Even-Odd conditions: odd and strictly increasing for even levels, and even and strictly decreasing for odd levels.

• Early Exit: If any level violates the conditions, immediately return False.

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

• Space Complexity: O(W), where W is the maximum width of the tree (maximum number of nodes at any level). In the worst case (complete binary tree), W can be approximately N.

Code

    from collections import deque
from typing import Optional

# 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

def isEvenOddTree(root: Optional[TreeNode]) -> bool:
    if not root:
        return True

    queue = deque([root])
    level = 0

    while queue:
        level_size = len(queue)
        prev_val = None

        for _ in range(level_size):
            node = queue.popleft()

            # Even level check
            if level % 2 == 0:
                if node.val % 2 == 0:
                    return False
                if prev_val is not None and node.val <= prev_val:
                    return False
            # Odd level check
            else:
                if node.val % 2 != 0:
                    return False
                if prev_val is not None and node.val >= prev_val:
                    return False

            prev_val = node.val

            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)

        level += 1

    return True