Description
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 index1
, their children are at level index2
, 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
Python3
# 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 isEvenOddTree(self, root: Optional[TreeNode]) -> bool:
isEven = True
queue = deque([root])
while queue:
N = len(queue)
last = None
for _ in range(N):
node = queue.popleft()
if isEven and (node.val % 2 == 0 or (last is not None and node.val <= last)):
return False
elif not isEven and (node.val % 2 == 1 or (last is not None and node.val >= last)):
return False
last = node.val
for leaf in filter(None, (node.left, node.right)):
queue.append(leaf)
isEven = not isEven
return True