二叉树 层序遍历算法

二叉树的层序遍历：
给定一个二叉树，层序遍历输出
输入：[3,9,20,null,null,15,7]
输出：[[3],[9],[20],[15],[7]]

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

class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        if not root:
            return []
        res = []
        queue = []
        queue.append(root)
        while queue:
            node = queue.pop(0)
            res.append(node.val)
            if node.left: queue.append(node.left)
            if node.right: queue.append(node.right)
        return res

二叉树的层序遍历（层层打印）：
输入：[3,9,20,null,null,15,7]
输出：[[3],[9,20],[15,7]]

class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        if not root:
            return []
        res = []
        queue = []
        queue.append(root)
        while queue:
            tmp = []
            for _ in range(len(queue)):
                node = queue.pop(0)
                tmp.append(node.val)
                if node.left: queue.append(node.left)
                if node.right: queue.append(node.right)
            res.append(tmp)    
        return res

二叉树层序遍历（层层打印）：
偶数层倒序

输入：[3,9,20,null,null,15,7]
输出：[[3],[20,9],[15,7]]

class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        if not root:
            return []
        res = []
        queue = []
        queue.append(root)
        while queue:
            tmp = []
            for _ in range(len(queue)):
                node = queue.pop(0)
                tmp.append(node.val)
                if node.left: queue.append(node.left)
                if node.right: queue.append(node.right)
            res.append(tmp[::-1] if len(queue)%2 else tmp)
        return res

参考链接：https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-ii-lcof/solution/mian-shi-ti-32-ii-cong-shang-dao-xia-da-yin-er-c-5/