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. 

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

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 。

• 广度优先遍历(分别判断奇数层和偶数层)

package leetcode typeTreeNodestruct { ValintLeft*TreeNodeRight*TreeNode } funcisEvenOddTree(root*TreeNode) bool { level:=0queue:= []*TreeNode{root} forlen(queue) !=0 { length:=len(queue) varnums []intfori:=0; i<length; i++ { node:=queue[i] ifnode.Left!=nil { queue=append(queue, node.Left) } ifnode.Right!=nil { queue=append(queue, node.Right) } nums=append(nums, node.Val) } iflevel%2==0 { if!even(nums) { returnfalse } } else { if!odd(nums) { returnfalse } } queue=queue[length:] level++ } returntrue } funcodd(nums []int) bool { cur:=nums[0] ifcur%2!=0 { returnfalse } for_, num:=rangenums[1:] { ifnum>=cur { returnfalse } ifnum%2!=0 { returnfalse } cur=num } returntrue } funceven(nums []int) bool { cur:=nums[0] ifcur%2==0 { returnfalse } for_, num:=rangenums[1:] { ifnum<=cur { returnfalse } ifnum%2==0 { returnfalse } cur=num } returntrue }