## DEV Community

Abhishek Chaudhary

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# 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
``````