Description
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 the 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, 105].
- 1 <= Node.val <= 10^6
分析
题目的意思是：判断一棵树是偶数奇数树，在偶数行，树是奇数，递增；在奇数行，树是偶数，递减。显然需要层序遍历，然后偶数行判断是否是递增，奇数行是否是递减就行了。用队列实现层序遍历。我看了一下别人的实现，跟我的思路差不多。
代码
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def isEvenOddTree(self, root: TreeNode) -> bool:
q=collections.deque()
q.append(root)
level=0
while(q):
n=len(q)
if(level%2==0):
prev=0
else:
prev=10**6+1
for i in range(n):
node=q.popleft()
if(level%2==0 and node.val%2!=1):
return False
elif(level%2==0 and prev>=node.val):
return False
if(level%2==1 and node.val%2!=0):
return False
elif(level%2==1 and prev<=node.val):
return False
if(node.left):
q.append(node.left)
if(node.right):
q.append(node.right)
prev=node.val
level+=1
return True