1609. Even Odd Tree

# 1609. 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:

``````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

## 题目大意 #

• 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

## 解题思路 #

• 广度优先遍历(分别判断奇数层和偶数层)

## 代码 #

``````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
}
``````

Apr 8, 2023