Even Odd Tree

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

SOLUTION:

# 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 levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []
        queue = [(root, 0)]
        levels = {}
        level = 0
        while len(queue) > 0:
            curr, level = queue.pop(0)
            levels[level] = levels.get(level, []) + [curr.val]
            if curr.left:
                queue.append((curr.left, level + 1))
            if curr.right:
                queue.append((curr.right, level + 1))
        return [levels[l] for l in range(level + 1)]

    def isOddAndIncreasing(self, arr):
        n = len(arr)
        for i in range(n - 1):
            if arr[i] >= arr[i + 1] or arr[i] % 2 == 0:
                return False
        if arr[-1] % 2 == 0:
            return False
        return True

    def isEvenAndDecreasing(self, arr):
        n = len(arr)
        for i in range(n - 1):
            if arr[i] <= arr[i + 1] or arr[i] % 2 == 1:
                return False
        if arr[-1] % 2 == 1:
            return False
        return True

    def isEvenOddTree(self, root: Optional[TreeNode]) -> bool:
        levels = self.levelOrder(root)
        for i, level in enumerate(levels):
            if i % 2 == 0 and not self.isOddAndIncreasing(level):
                return False
            elif i % 2 == 1 and not self.isEvenAndDecreasing(level):
                return False
        return True