103. 二叉树的锯齿形层次遍历

思路：

层次遍历：

下面这中方式复杂化了：变动了结点

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def zigzagLevelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if root is None:
            return []
        results = []
        queue = [root]
        count = 1  # 记录层数
        while queue:
            count += 1
            result = [] # 记录下一层的值
            now_queue = []
            nums = len(queue)  # 记录每层的结点个数
            for i in range(nums):
                node = queue.pop(0)
                result.append(node.val)
                left = node.left
                right = node.right
                if count % 2 == 0:
                    if left:
                        now_queue.append(left)
                    if right:
                        now_queue.append(right)
                else:
                    if right:
                        now_queue.append(right)
                    if left:
                        now_queue.append(left)
            queue = now_queue[-1::-1]
            results.append(result)
        return results

其实很简单，就是层次遍历，只不过是将奇数层的值翻转一下！！！

class Solution(object):
    def zigzagLevelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root:
            return []
        queue = [root]
        results = []
        count = 0
        while queue:
            count += 1
            nums = len(queue)
            result = []
            for i in range(nums):
                node = queue.pop(0)
                left = node.left
                result.append(node.val)
                right = node.right
                if left:
                    queue.append(left)
                if right:
                    queue.append(right)
            if count % 2 == 0:
                result = result[::-1]
            results.append(result)
        return results