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:

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.

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 #

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
}

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