【leetcode】1609. Even Odd Tree

题目如下：

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 <= 106

解题思路：题目不难，就是一个二叉树的层序遍历。由于每一层的所有节点的值是递增或者递减，因此遍历过程中要记录每一层上一个遍历的节点的值，用以和当前节点的值比较。

代码如下：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    dic = {}
    res = True
    def recursive(self,node,level):
        if self.res == False:
            return
        elif level % 2 == 0:
            if node.val % 2 == 0:
                self.res = False
                return
            elif self.dic.has_key(level) == False:
                self.dic[level] = node.val
                if node.left != None:
                    self.recursive(node.left,level+1)
                if node.right != None:
                    self.recursive(node.right, level + 1)
            elif self.dic[level] >= node.val:
                self.res = False
                return
            else:
                self.dic[level] = node.val
                if node.left != None:
                    self.recursive(node.left,level+1)
                if node.right != None:
                    self.recursive(node.right, level + 1)
        elif level % 2 == 1:
            if node.val % 2 == 1:
                self.res = False
                return
            elif self.dic.has_key(level) == False:
                self.dic[level] = node.val
                if node.left != None:
                    self.recursive(node.left,level+1)
                if node.right != None:
                    self.recursive(node.right, level + 1)
            elif self.dic[level] <= node.val:
                self.res = False
                return
            else:
                self.dic[level] = node.val
                if node.left != None:
                    self.recursive(node.left,level+1)
                if node.right != None:
                    self.recursive(node.right, level + 1)
    def isEvenOddTree(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        self.res = True
        self.dic = {}
        self.recursive(root,0)
        return self.res