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150 lines
3.7 KiB
Markdown
150 lines
3.7 KiB
Markdown
# [1609. Even Odd Tree](https://leetcode.com/problems/even-odd-tree/)
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## 题目
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A binary tree is named Even-Odd if it meets the following conditions:
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- 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.
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- For every even-indexed level, all nodes at the level have odd integer values in strictly increasing order (from left to right).
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- For every odd-indexed level, all nodes at the level have even integer values in strictly decreasing order (from left to right).
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Given the root of a binary tree, return true if the binary tree is Even-Odd, otherwise return false.
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**Example 1**:
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Input: root = [1,10,4,3,null,7,9,12,8,6,null,null,2]
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Output: true
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Explanation: The node values on each level are:
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Level 0: [1]
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Level 1: [10,4]
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Level 2: [3,7,9]
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Level 3: [12,8,6,2]
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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.
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**Example 2**:
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Input: root = [5,4,2,3,3,7]
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Output: false
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Explanation: The node values on each level are:
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Level 0: [5]
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Level 1: [4,2]
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Level 2: [3,3,7]
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Node values in level 2 must be in strictly increasing order, so the tree is not Even-Odd.
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**Example 3**:
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Input: root = [5,9,1,3,5,7]
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Output: false
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Explanation: Node values in the level 1 should be even integers.
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**Example 4**:
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Input: root = [1]
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Output: true
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**Example 5**:
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Input: root = [11,8,6,1,3,9,11,30,20,18,16,12,10,4,2,17]
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Output: True
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**Constraints:**
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- The number of nodes in the tree is in the range [1, 100000].
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- 1 <= Node.val <= 1000000
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## 题目大意
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如果一棵二叉树满足下述几个条件，则可以称为 奇偶树 ：
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- 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。
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- 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
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- 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
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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 。
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## 解题思路
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- 广度优先遍历(分别判断奇数层和偶数层)
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## 代码
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```go
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package leetcode
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type TreeNode struct {
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Val int
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Left *TreeNode
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Right *TreeNode
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}
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func isEvenOddTree(root *TreeNode) bool {
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level := 0
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queue := []*TreeNode{root}
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for len(queue) != 0 {
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length := len(queue)
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var nums []int
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for i := 0; i < length; i++ {
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node := queue[i]
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if node.Left != nil {
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queue = append(queue, node.Left)
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}
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if node.Right != nil {
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queue = append(queue, node.Right)
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}
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nums = append(nums, node.Val)
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}
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if level%2 == 0 {
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if !even(nums) {
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return false
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}
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} else {
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if !odd(nums) {
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return false
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}
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}
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queue = queue[length:]
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level++
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}
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return true
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}
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func odd(nums []int) bool {
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cur := nums[0]
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if cur%2 != 0 {
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return false
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}
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for _, num := range nums[1:] {
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if num >= cur {
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return false
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}
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if num%2 != 0 {
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return false
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}
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cur = num
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}
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return true
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}
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func even(nums []int) bool {
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cur := nums[0]
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if cur%2 == 0 {
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return false
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}
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for _, num := range nums[1:] {
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if num <= cur {
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return false
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}
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if num%2 == 0 {
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return false
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}
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cur = num
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}
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return true
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}
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``` |