1609. Even Odd Tree.py

'''
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 the 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.

Example 4:
Input: root = [1]
Output: true

Example 5:
Input: root = [11,8,6,1,3,9,11,30,20,18,16,12,10,4,2,17]
Output: true
 
Constraints:
The number of nodes in the tree is in the range [1, 105].
1 <= Node.val <= 106
'''

from collections import deque
class Solution:
    def isEvenOddTree(self, root):
        q = deque()
        q.append(root)
        level = 0
        while(1):
            if valid(q, level):
                new = []
                while(q):
                    n = q.popleft()
                    if n.left:
                        new.append(n.left)
                    if n.right:
                        new.append(n.right)
                if len(new)==0:
                    break
                q = deque(new)
            else:
                return False

            level+=1
        
        return True

def valid(arr, l):
    if l%2:
        if arr[0].val%2==0:
            for i in range(1,len(arr)):
                if arr[i].val>=arr[i-1].val or arr[i].val%2:
                    return False
            return True
        else:
            return False
    else:
        if arr[0].val%2:
            for i in range(1,len(arr)):
                if arr[i].val<=arr[i-1].val or arr[i].val%2==0:
                    return False
            return True
        else:
            return False