# [1609. Even Odd Tree](https://leetcode.com/problems/even-odd-tree/)

## 题目

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](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](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](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.

**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, 100000]. 
- 1 <= Node.val <= 1000000

## 题目大意

如果一棵二叉树满足下述几个条件，则可以称为 奇偶树 ：

- 二叉树根节点所在层下标为 0 ，根的子节点所在层下标为 1 ，根的孙节点所在层下标为 2 ，依此类推。
- 偶数下标 层上的所有节点的值都是 奇 整数，从左到右按顺序 严格递增
- 奇数下标 层上的所有节点的值都是 偶 整数，从左到右按顺序 严格递减

给你二叉树的根节点，如果二叉树为 奇偶树 ，则返回 true ，否则返回 false 。

## 解题思路

- 广度优先遍历(分别判断奇数层和偶数层)

## 代码

```go
package leetcode

type TreeNode struct {
	Val int
	Left *TreeNode
	Right *TreeNode
}

func isEvenOddTree(root *TreeNode) bool {
	level := 0
	queue := []*TreeNode{root}
	for len(queue) != 0 {
		length := len(queue)
		var nums []int
		for i := 0; i < length; i++ {
			node := queue[i]
			if node.Left != nil {
				queue = append(queue, node.Left)
			}
			if node.Right != nil {
				queue = append(queue, node.Right)
			}
			nums = append(nums, node.Val)
		}
		if level%2 == 0 {
			if !even(nums) {
				return false
			}
		} else {
			if !odd(nums) {
				return false
			}
		}
		queue = queue[length:]
		level++
	}
	return true
}

func odd(nums []int) bool {
	cur := nums[0]
	if cur%2 != 0 {
		return false
	}
	for _, num := range nums[1:] {
		if num >= cur {
			return false
		}
		if num%2 != 0 {
			return false
		}
		cur = num
	}
	return true
}

func even(nums []int) bool {
	cur := nums[0]
	if cur%2 == 0 {
		return false
	}
	for _, num := range nums[1:] {
		if num <= cur {
			return false
		}
		if num%2 == 0 {
			return false
		}
		cur = num
	}
	return true
}
```