Even Odd Tree
Tags: leetcode, dsa, theabbie
Published: 2022-06-07 10:00:48

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:**

![](https://assets.leetcode.com/uploads/2020/09/15/sample_1_1966.png)

**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:**

![](https://assets.leetcode.com/uploads/2020/09/15/sample_2_1966.png)

**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:**

![](https://assets.leetcode.com/uploads/2020/09/22/sample_1_333_1966.png)

**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